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Full text of "Geometrical drawing and design"

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GEOMETRICAL DRAWING AND DESIGN. 



0m, 



MACMILLAN AND CO., Limited 

LONDON • BOMBAY • CALCUTTA 
MELBOURNE 

THE MACMILLAN COMPANY 

NEW YORK • BOSTON • CHICAGO 
DALLAS • SAN FRAKCLSCO 

THE MACMILLAN CO. OF CANADA, Ltd, 

TORONTO 



GEOMETRICAL 
DRAWING AND DESIGN 



BY 

J. HUMPHREY SPANTON 

GOLD MEDALLIST, ROYAL ACADEMY OF ARTS, LONDON 

INSTRUCTOR IN DRAWING TO NAVAL CADETS IN H.M.S. "hRITANNIa' 

AUTHOR OF "complete PERSPECTIVE COURSE*' 



ADAPTED TO THE REQUIREMENTS OF 
THE BOARD OF EDUCATION 



MAC^riLLAN AND CO., LIMITED 

ST. MARTIN'S STREET, LONDON 

1913 



'S'SIOO 



1 



COPYRIGHT. 

First Eflition 100?. 

Repi-intcd ]004; with alterations and additions 1006. 

Reprinted 1009, 1011, with additions 1013. 



Me. 

Si 



GLASGOW : PRINTKD AT THE UNIVERSITV PRESS 
BY ROBERT IMACt.EHOSE AND CO. LTD. 



■Mi' 



PREFACE. 



A COURSE of geometrical drawing or practical geometry pro- 
vides a valuable preliminary training for so many handicrafts 
and professions that it must be regarded as essential to all 
students whose work is to be adapted to modern requirements. 
The Engineer, the Architect, the Soldier, and the Statistician, 
all have recourse to the assistance of practical geometry to solve 
their problems or to explain their methods. Every day the 
graphic treatment of subjects is finding its application in new 
directions ; and to be able to delineate the proportions of any 
scheme places in the hand an invaluable tool for the execution 
of work of practical value. 

The Author's complete geometrical course has now for some 
time been widely used for the advanced parts of the subject; 
such as Projection, Section and Interpenetration of Solids. 
Hence the publication of its simpler parts as an introduction 
to Design would seem likely to meet with favour. 

The geometrical basis of ornamentation is the rational one, 
though youth and fancy might condemn it as a chaining of 
Pegasus and the curbing of imagination. It is no doubt the 
height of art to conceal the scheme of treatment and delight 
the eye with novel suggestions of development. But underlying 
all is the demand of Nature for order and rhythm, such as 
can be obtained by a study of geometrical figures and designs. 

The course of work prescribed in Geometrical Drawing (Art) 
by the Board of Education aims at giving students the ability 
to construct ordinary geometrical figures, and the power to 



PREFACE. 



apply these figures as the basis of ornamental and decorative 
work. In the preparation of the present volume these intentions 
have been borne in mind, but the scope of the work has not 
been limited by the syllabus of the subject. The greater 
part of the book contains an elementary course of constructive 
geometry suitable for all students, and the sections which show 
the applications of geometrical constructions to design, while 
of especial importance to students of appHed art, are not 
without interest to students of science. Moreover, familiarity 
with the constructions in the early part of the book provides 
the best introduction a pupil could have to the study of formal 
geometry. 

The Editor's thanks are due to Prof Thos. C. Simmonds, 
the Headmaster of the Municipal School of Art, Derby, for 
his valuable advice, and to Mr. E. E. Clark, his assistant, by 
whom the drawings for the sections on Design have been 
furnished. Acknowledgment must also be made of assistance 
rendered by Prof R. A Gregory in arranging the book and 
seeing it through the press. 



CONTENTS. 

PAGE 

Drawing Instruments and Materials. - - . . - i 

General Directions, --------- 3 

CHAPTER I. 
Geometrical Definitions and General Properties, ... 6 

PLANE GEOMETRY. 

CHAPTER II. 
Lines, Triangles, Quadrilaterals, Convergent Lines, and Circles, 15 

CHAPTER III. 
General Information concerning Polygons, - - - - 30 
General Methods for constructing Polygons, - - - - 32 

CHAPTER IV. 
Inscribed and Described Figures, 40 

CHAPTER V. 
Foiled Figures, 53 



viii CONTENTS 

CHAPTER VI. 

PAGE 

Tangents a- ' tangential Arcs, 57 

CHAPTER VIL 

Proportional Lines, - - - 72 

CHAPTER Vni. 

Construction of Plain Scales, - - - - - - - 81 

,, ,, Comparative Scales, 84 

,, ,, Diagonal Scales, 86 

,, ,, Proportional Scales, 89 

CHAPTER IX. 

The Use of the Protractor, ------- 92 

The Use of the Sector, - - 94 

CHAPTER X. 

Construction of Similar Figures, 101 

Principles of Similitude, or the Enlargement and Reduction of 

Plane Figures, 103 

CHAPTER XL 

The Conic Sections — 107 

Practical Method for drawing the Conic Sections, - - 107 

Mechanical Methods for drawing the Ellipse, - - - 108 

The Ellipse, - - 108 

The Parabola, 112 

The Hyperbola, - - - - - - - - 113 

Mechanical Methods for drawing the Parabola and Hyper- 
bola, 114 

Cycloidal Curves, 116 

Spirals, - - - - - - - - - -121 



CONTENTS ix 

SOLID -GEOMETRY. 

CHAPTER XII. 

Page 

itroduction, 126 

CHAPTER XIII. 

Simple Solids in given Positions to Scale, - - - - 132 

The Regular Solids, 142 

Octagonal Pyramids, Cones, Cylinders, and Spheres, - - 146 

CHAPTER XIV. 

\ rthographic Projection, - - - 157 

,, Lines, 158 

,, Planes, 162 

CHAPTER XV. 

Sections of Solids, Construction of Sectional Areas, - - 169 

,, ,, a Cone, - - - - - - - - '75 

„ ,, a Cylinder, - - I79 

,, ,, a Sphere, 180 



DESIGN. 

CHAPTER XVI. 

Construction Lines on which Patterns are arranged, - - - 184 

Units of Pattern, . . - 193 

Spacing of Walls and other Surfaces, 196 

Bands and Borders, - - - - - - - - 197 

Defined areas, - 202 



CONTENTS 



CHAPTER XVII. 

Lettering 

Shields, 

Diaper, Chequer, Spot, Powder, ... - 

CHAPTER XVIIL 
Arch Forms and Tracery, . . . = = 
Greek and Roman Mouldings, - - - - = 
Gothic Mouldings, - - ..... 
Gothic Piers, - - 

Miscellaneous Exercises, 

Examination Papers in Geometrical Drawing, 



PART I. 
PRACTICAL PLANE GEOMETRY. 

INTRODUCTION. 

Drawing Instruments and Materials. 

The Drawing-board. — A very convenient size to use for 
ordinary purposes is half Imperial (23'' x 16") ; it should be 
made of well-seasoned yellow pine, with the edges true and at 
right angles to each other. 

The Tee-square.— This is a ruler with a cross piece or stock 
at the end : it is like the letter T in shape, hence its name. 
The blade should be screwed on to the stock, and not mortised 
into it, so as to allow of the set-squares being used up to the 
extreme margin of the paper, as illustrated in Fig. i. By 
keeping the stock of the tee-square pressed closely against the 
edge of the drawing-board, we are enabled to draw lines parallel 
to each other. 

Set-squares. — These are right-angled triangles made with 
given fixed angles out of thin pieces of wood or ebonite : the 
latter is preferable, as it is not liable to warp. The most useful 
angles are those of 45° and 60°. 

French Curves. — These are thin pieces of wood cut into a 
variety of curves. They are used for drawing fair curves, that 
are not arcs of circles, through a succession of points : the 
cycloidal curves, for instance. 

Scale. — A plain scale about 6 inches long, divided into inches, 
with sub-divisions of eighths on one edge and tenths on the 
other. 

CI A 



2 geomp:trical drawing and design. 

Pencils. — Two degrees of hardness should be used : HH for 
drawing in the construction, and F for drawing in the result 
with a firmer line. 

Drawing-paper.— This should have a hard smooth surface. 
Whatman's "hot-pressed" is the best for fine work ; but if the 
drawing has to be coloured, a damp sponge should first be 
drawn across the surface, to remove the gloss. Cartridge-paper 
of good quality is suitable for ordinary work. 

The most convenient size is "Imperial" (30" x 22";, which can 
be cut to half, or quarter Imperial, as desired. 

Drawing-pins. — These should have short fine points, so as 
not to make large holes in the drawing-board. 

Dividers. — These are used for setting off distances or 
dividing lines. There is a special kind made, called "hair- 
dividers," one leg of which can be adjusted by means of a 
spring and screw : these are very useful for dividing lines, etc. 

Ruling-pen. — This is used for inking in lines, the thickness 
of which is regulated by a screw. Some are made in which the 
nib that works against the ruler is of an extra thickness of metal : 
this is to prevent the nibs from closing when the pen is pressed 
against the ruler. 

Bow-pencil. — This is a small pair of compasses with one leg 
constructed to hold a pencil, used for drawing circles and arcs. 

Bow-pen. — This is a similar instrument to a bow-pencil, but 
with a ruling pen for one of its legs instead of a pencil ; it is 
used for inking in circles and arcs. 

Note. — Both the bov/-pencil and bow-pen should have hinged 
legs ; because, when a number of circles are drawn from the 
same centre, they are likely to make a large hole in the paper, 
unless the leg used for the centre is kept perpendicular to the 
paper. It is also necessary to have the pen-leg as upright as 
possible, otherwise it has a tendency to draw uneven lines. 

Compasses.— Full-sized compasses are suppHed, with inter- 
changeable arms, divider, pen and pencil, opening to 12 inches. 

Indian ink should be used for inking in a drawing. It has 
several advantages over common ink : it dries quickly ; it does 
not corrode the ruling-pen ; and the lines can be coloured over 
without their running, if a waterproof quality is used. 

The most convenient is the liquid Indian ink, sold in bottles, 



PRACTICAL PLANE GEOMETRY 



as it is always ready for use. The ruling-pen should be filled 
with Indian ink by means of an ordinary steel nib. If the cake 
Indian ink is used, after rubbing it in a saucer, a piece of thin 
whalebone should be used for filling the ruling-pen. 

General Directions. 

Keep all instruments perfectly clean : do not leave ink to 
dry in the ruHng-pen. 

In using dividers avoid, as much as possible, making holes 
through the paper. 

The paper should be firmly fixed to the drawing-board by 
a drawing-pin at each corner, well pressed down. Do not stick 
pins in the middle of the board, because the points of the 
dividers are liable to slip into them and make unsightly holes 
in the paper. 

A pencil sharpened to what is called a " chisel-point " is 
generally used for drawing lines ; it has the advantage of retain- 
ing its point longer, but a nicely-pointed pencil is better for 
neat work, as it enables you to see the commencement and 
termination of a line more easily. 




Fig. I. 

Always rule a line from left to right, and slope the pencil 
slightly towards the direction in which it is moving ; if this is 
done, there is less chance of indenting the paper, which should 
always be avoided. 



4 GEOMETRICAL DRAWING AND DESIGN. 

Having determined the extent of a line, always rub out the 
superfluous length ; this will prevent unnecessary complication. 

Avoid using India-rubber more than is necessary, as it tends 
to injure the surface of the paper. After inkmg in a drawings 
use stale bread in preference to India-rubber for cleaning it up. 

The tee-square should be used for drawing horizontal lines 
only (Fig. i) ; the perpendicular lines should be drawn by 
the set-squares. If this is done, it is immaterial whether the 
edges of the drawing-board are at right angles, because it will 
only be necessary to use one of its edges. 

For drawing parallel lines that are neither horizontal nor 
perpendicular, hold one set-square firmly pressed upon the 
paper and slide the other along its edge (Fig. 2). Geometrical 
drawing can be greatly facilitated by the proper use of set- 
squares, so it is advisable to practise their use. 




Fig. 2. 

When a problem contains many arcs of circles, it is advisable 
to connect each arc with its corresponding centre. Enclose the 
centre in a small circle ; draw a dotted line to the arc, termi- 
nated by an arrow-head (Fig. 3). 

In drawing intersecting arcs for bisecting lines, etc., the arcs 
should not intersect each other too obtusely or too acutely : the 
nearer the angle between the arcs approaches 90° the easier it 
will be to ascertain the exact point required. 



PRACTICAL TLANE GEOMETRY. 



In joining two points by a line, first place the point of the 
pencil on one point, then place the edge of the ruler against it, 
and adjust the ruler till its edge coincides with the other point. 




All the problems should be drawn larger than shown. 

Great accuracy is required in drawing the various problems. 
Every effort should be made to ensure neatness and precision 
in the work. 

All arcs should be inked in first, as it is easier to join a hne to 
an arc than an arc to a line. 



CHAPTER I. 
GEOMETRICAL DEFINITIONS, 

A point simply marks a position ; it is supposed to have no 
magnitude. 

A line has length without breadth or thickness. The ex- 
tremities and intersections of lines are points. A straight line 
or right line is one that is in the same direction throughout 
its length, and is the shortest that can be drawn between two 
points. To produce a line is to lengthen it. 

A plane is a flat even surface ; it has length and breadth 
only. The intersections of planes are straight lines. 

Parallel lines are straight lines in the 

same plane, and at equal distances apart 

throughout their entire length ; if pro- " 

duced they would never meet (Fig. 4), ^'^- 4- 

A circle is a plane figure bounded by a curved line, such that 
all straight lines drawn to it from a 
certain point are equal. This point is 
called the centre, and the curved line 
is called the circumference of the circle. 

A straight line drawn from the centre 
to the circumference, as ce or cd (Fig. 5) 
is called a radius. A straight line drawn 
through the centre, and terminated at 
both ends by the circumference, as ab^ 
is called a diameter. A semicircle is 
half a circle, as adb. A quadrant is a quarter of a circle, as adc. 




GEOMETRICAL DEFINITIONS. 



An arc is any portion of ihe circumference of a circle, as abc 
(Fig. 6). A chord is a straight line join- 
ing the extremities of an arc, as ac. A 
segment is the space enclosed by the 
arc and its chord, as f. A sector is the 
space enclosed by two radii and the 
arc between them, as g. A tangent is 
line touching the circumference, as (ie ; 
it is always at right angles to the radius 
of the circle at the point of contact. 





Fig. 9. 



An ordinate is a line drawn from a point m 
a curve perpendicular to the diameter, as dotted 
line in Fig. 7. 



An abscissa is the part of the diameter cut 
off by the ordinate, as dotted line in Fig. 8. 



An angle is the inclination of two 
straight lines meeting in a point. This 
point is called the vertex of the angle, 
as a (Fig. 9). The angle here shown 
would be called either dac or cad. 

If two adjacent angles made by two straight lines at the 
point where they meet be equal, as f^ca and deb (Fig. 5), each 
of these angles is called a right angle, and either of the straight 
lines may be said to be perpendicular to the other. 

A right angle is supposed to be divided into 90 equal parts, 
each of which is called a degree. A degree is expressed in 
writing by a small circle placed over the last figure of the 
numerals denoting the number of degrees — thus 36° means 
thirty-six degrees. 

The circumference of a circle is supposed to be divided into 
360 equal arcs, each of which subtends an angle of 1° at the 
centre. Sometimes this arc is itself loosely termed a degree. 

An angle containing more than 90° is called an obtuse angle, 
as ecd (Fig. 5) ; while an angle containing less than 90° is called 
an acute angle, as ace. 



8 



GEOMETRICAL DRAWING AND DESIGN. 



A line is said to be perpendicular to a plane when it is at 
right angles to any straight line in that plane which meets it. 
Concentric circles have the same centre. 

Triangles. 

Triang-ies are closed figures contained by three straight lines. 

A triangle which has all its sides equal is 
called equilateral (Fig. lo). 

N.B. — Such a triangle will always have its 
three angles equal, and therefore will also be 
equiangular. 



A triangle which has two sides (and there- 
fore two angles) equal is called isosceles (Fig. 1 1) 
{isos^ equal ; skelos, a leg). 



A scalene triangle has none of its sides equal. 

A triangle which has a right angle is called 
right-angled (Fig. 12). The side opposite to the 
right angle, as ab, is called the hypotenuse {hypo, 
under ; te?iein, to stretch). 



A triangle which has an obtuse angle 
is called obtuse-angled (Fig. 1 3). 



A triangle which has three acute angles 
is called acute angled (Fig. 14). 

Fig. 14. 

The base of a triangle is its lowest side, as al? (Fig. 10). 

The vertex is the point opposite the base, as c (Fig. 10). 

The altitude or perpendicular height is a line drawn from 
the vertex at right angles to its base, as cd {¥\^. 10). 

The median is a line drawn from the veitex to the middle 
point of the base. 



GEOMETRICAL DEFINITIONS. 



Quadrilateral Figures. 

Quadrilateral figures are such as are bounded by four straight 
lines. 

A quadrilateral figure whose 
opposite sides are parallel is 
called a parallelogram (Fig. 15). 
N.B. — The opposite sides and 

angles of parallelograms are ^ -^d 

equal. fig- ^s- 




A parallelogram whose angles are 
right angles is called a rectangle 
(Fig. 16) or oblong. 



A rectangle which has its sides equal 
called a square (Fig. 1 7). 



A parallelogram whose angles 
are not right angles is called 
a rhombus, if its sides are all 
equal (Fig. 18), or a rhomboid if 
the opposite sides alone are 
equal — Gk. rhombos^ from rhein- 
bein^ to twirl, from some likeness 
to a spindle. 

All other quadrilaterals are called trapeziums. 

A line joining two opposite angles of a quadrilateral figure is 
called a diagonal, as the dotted line ab (Fig. 1 5). 




Polygons. 

A polygon is a plane figure which has more than four angles. 
A polygon which is both equilateral and equiangular is called 
regular. 



10 GEOMETRICAL DRAWING AND DESIGN. 

A polygon of five sides is called a pentagon. 
„ „ six „ „ hexagon. 

„ „ seven „ „ heptagon. 

„ „ eight „ „ an octagon, etc. 

Solids. 

A solid has length, breadth, and thickness. A solid bounded 
wholly by planes is called a polyhedron {poly^ many ; hedra^ a 
side). 

A solid bounded by six planes or faces, whereof the opposite 
ones are parallel, is called a parallelepiped {pa?-aUelos, parallel ; 
and epipedon^ a plane). 

A parallelepiped whose angles are all right angles is called a 
rectangular parallelepiped or orthohedron {ort/ios, right ; and hedra^ 
a side). 

An orthohedron with six equal faces is called a cube {kudos, a 
die). 

A polyhedron, all but one of whose faces meet in a point, is 
called a pyramid (Gk. pyrainis, a pyramid). 

Pyramids are often named, after the shape of their bases, 
triangular, square, etc. 

A polyhedron, all but two of whose faces are parallel to one 
straight line, is called a prism (Gk. prisma, irom. prizein, to saw, 
a portion sawn off). 

If the ends of a prism are at right angles to the straight line 
to which the other faces are parallel it is called a right prism. 

Prisms are often named, after the shape of their ends, 
triangular, hexagonal, etc. 

A cylinder is a solid described by the revolution of a rect- 
angle about one of its sides which remains fixed. This fixed 
line is called the axis of the cylinder. 

A right circular cone — generally spoken of simply as a cone 
— is a solid described by the revolution of a right-angled triangle 
about one of the sides containing the right angle, which side 
remains fixed. This fixed line is called the axis of the cone ; 
the base is a circle, and the point opposite the base is called 
the vertex. 

A solid bounded by a closed surface, such that all straight 



GEOMETRICAL DEFINITIONS. 



lines drawn to it from a certain point are equal, is called a 
sphere (Gk. sphatra, a ball). 

The point referred to is called the centre of the sphere. 



Technical Definitions. 

A plane parallel to the ground, or, more strictly speaking, 
parallel to the surface of still water, is called a horizontal plane. 

A vertical plane is a plane at right angles to a horizontal 
plane. 

A horizontal line is a line parallel to a horizontal plane. 

A vertical line is a line at right angles to a horizontal plane. 





Fig. 19. 

The plan of an object is the tracing made on a horizontal plane 
by the foot of a vertical line, which moves so as to pass succes- 
sively through the various points and outlines of the object, as 
A and B (Fig. 19). 



12 



GEOMETRICAL DRAWING AND DESIGN. 



An elevation of an object is the tracing made on a vertical 
plane by the end of a horizontal line, at right angles to the 
vertical plane, which moves so as to pass successively through 
the various points and outlines of the object, as A' and B' 
(Fig. 19). 

N.B. — A is an object parallel to the vertical plane, and B an 
object inclined to it. 



General Properties of some of the Figures 
already described. 

If two lines cross each other the opposite angles are always 
equal. The angle acb is 
equal to the angle dee, and 
the angle aed is equal to the 
angle bee (Fig. 20). ^" 

The two adjacent angles 
are equal to two right angles ; 
the angles aed and aeb^ for 
instance, as well as the angles bee and 



Fig. 20. 



:d. 



Triangles.— The three angles of a triangle con- 
tain together 180°, or two right angles ; so if two 
angles are given, the third angle can always be 
found. F^or example, if one angle is 70° and the 
other 30°, the remaining angle must be 80° (Fig. 21). 
70° + 30°= 100°. 
1 80°- 1 00° = 80°. 



The exterior angle of a 
triangle is equal to the two 
opposite interior angles. 
The angle abe is equal to the 
angles bed and edb together ; 
in the same way the angle 
ede is equal to the two angles ^ 
bed and ebd (Fig, 22). 




b d e 

Fig. 22. 

If we multiply the base by half the altitude, we get the area 
of a triangle ; or half its base by its altitude will give us the 
same result. 



GEOMETRICAL DEFINITIONS. 



!3 



Triangles of equal bases drawn between parallel lines are 
equal in area, and lines drawn parallel to their bases at equal 
heights are equal in length, as the dotted lines shown (Fig. 23). 




Fig. 23. 



If we bisect two sides of a triangle and join 
the points of bisection, we get a line that is 
always parallel to the third side (Fig. 24). 



Quadrilaterals. — If we bisect the four 
sides of a quadrilateral figure and join 
the points, it will always give us a 
parallelogram, as shown by dotted 
lines. The reason for this will be 
apparent by the principle shown in the 
preceding figure if we draw a diagonal 
of the quadrilateral, so as to form two 
triangles (Fig. 25). 




Fig. 25. 



Parallelograms drawn between parallel lines on equal bases 
are always equal in area, and parallel lines drawn at equal 
heights are always equal to each other and to the bases, as 
shown by dotted lines (Fig. 26). 




Fig, 26. 



14 



GEOMETRICAL DRAWING AND DESIGN. 



Semicircles. — Any two lines drawn 
from the extremities of the diameter, 
to a point on the circumference of a 
semicircle, will form a right angle. 




Fig. 27. 



EXERCISES. 

Note. — Feet are represented by one dash {'), and inches by two 
dashes (") ; 3 feet 6 inches would be written thus — 3' 6". 

1. Draw lines of the following lengths: 3", 45", 2|", ig", 2.25", 
3.50", 1.75". 

2. Draw an acute angle, and an obtuse angle. 

3. Draw the following triangles, viz. equilateral, scalene, isosceles, 
obtuse-angled, right-angled, and acute-angled. 

4. Draw a right-angled triangle, and write the following names to 
its different parts, viz. hypotenuse, vertex, base, median, and altitude. 

5. Draw the following figures, viz. rectangle, rhombus, square, 
rhomboid, trapezium, and parallelogram. 

6. Draw a circle, and name the different parts, viz. sector, radius, 
chord, arc, diameter, segment, and tangent. 



CHAPTER 11. 

PROBLEMS ON LINES, TRIANGLES, QUADRILATERALS, 
CONVERGENT LINES, AND CIRCLES. 



Lines. 



1. To bisect a given straight line 
AB. 

From A and B as centres, with 
any radius greater than half the 
hne, describe arcs cutting each 
other in C and D ; join CD. The 
straight hne CD will bisect AB in 
E. Also the line CD will be per- 
pendicular to AB. 



2. To bisect a given arc AB. 

Proceed in the same way as in 
Problem i, using the extremities 
of the arc as centres. The arc 
AB is bisected at E. 




i6 GEOMETRICAL DRAWING AND DESIGN. 

3. To draw a line parallel to a given line AB through a given 
point C. 

p ^ ^ Take any point D in line 

,''p -^ AB, not opposite the point 

\ \ C ; with D as centre, and DC 

\ V as radius, describe an arc 

\ « cutting AB in E, and from 

\ 15 £_'. C as centre, with the same 

c_ ' ■" B radius, draw another arc DF ; 

Fig. 30- set off the length EC on DF ; 

a line drawn through CF will be parallel to AB. 



4. To draw a line parallel to a given line AB at a given distance 
from it. 

p r. Let the length of the line C 

"^ '' "" represent the given distance. 

Take any points D and E 

in line AB as centres, and 

with C as radius describe 

Ad E ^ arcs as shown ; draw the line 

FG as a tangent to these 
arcs. FG will be parallel to 



Fig. 31- AB. 



5. From a point C in a given line AB, to draw a line 
perpendicular to AB. 



F 



c 

Fig. 32. 



At the point C, with any 
radius, describe arcs cutting 
AB in D and E ; with D 
and E as centres, and with 
any radius, draw arcs inter- 
secting at F ; join FC, 

c < which will be perpendicular 

— ! r, to AB. 



PROBLEMS. 



6. To draw a perpendicular to AB from a point at, or near, the end 
of the given line. 



With B as centre, and with 
any radius BC, draw the arc 
CDE ; and with the same 
radius, starting at C, set off 
points D and E ; with each of 
these two points as centres 
and with any radius, draw arcs 
cutting each other at F ; join 
FB, which will be perpendicular 
to AB. 



F^ 






Fig- 33- 



7. To draw a line perpendicular to a given line, from a point 

which is without the line. 

C 

Let AB be the given line 
and C the point. 

With C as centre, and with 
any radius greater than CD, 

draw arcs cutting the line i— ___ 

AB in E and F ; from these ^ ^^ ~ 
points as centres, with any 
radius, describe arcs inter- 
secting at G ; join CG, which 
will be perpendicular to AB. 

Fig. 34. 

8. To draw a perpendicular to AB from a point opposite, or nearly- 

opposite, to one end of the Une. 

r 
Let C be the given point. 
Take any point D in AB, not 
opposite the point C ; join 
CD and bisect it in E ; with 
E as centre, and EC as radius, 
draw the semi-circle CFD ; 
join CF, which will be per- 
pendicular to AB. 



zy 



Fig. 35- 



GEOMETRICAL DRAWING AND DESIGN. 



To divide a given line AB into any number of equal parts. 
Take five for example. 



.D 



B 



Fig. 36. 

into five equal parts. 



F'rom one extremity A, 

draw AD, at any angle to 

AB ; and from B, draw BC, 

parallel to AD. With any 

convenient radius, set off 

along AD, commencing at A, 

^ , K'" ' four parts (the number of 

parts, less one, into which it 

is required to divide the given line), repeat 

the same operation on BC, commencing at 

B, with same radius ; join the points as 

shown, and the given line AB will be divided 



10. 

c 

/A 



Another Method. 

Draw AC at any angle to 
AB ; AC may be of any con- 
venient length. With any 
radius, mark off along AC 
the number of equal parts re- 
\ quired ; join the last division 

\ C with B ; draw lines from 

__i all the other points parallel to 

B CB, till they meet AB, which 
will be divided as required. 



Fig. 37- 

m. From a given point B, in a given line AB, to construct an 



angle equal to a given angle C. 




From point C of the given 
angle as centre, and with 
any radius, draw the arc 
EF : and with the same 
radius, with B as centre, 
draw the arc GH ; take the 
length of the arc EF, and 
set it off on GH ; draw the 
line BD through H. Then 
the angle GBH will be equal 
to the given angle C. 



PROBLEMS. 



12. 



To bisect a given angle 
ABC. 




From B as centre, and with 
any radius, draw the arc AC ; 
from A and C as centres, with 
any radius, draw the arcs inter- 
secting at D ; join DB, which 
will bisect the angle ABC. 



13. To trisect a right angle. 

Let ABC be the right angle. 
From B as centre, and with any 
radius, draw the arc AC ; with 
the same radius, and A and C as 
centres, set off points E and D; 
join EB and DB, which will 
trisect the right angle ABC. 

Fig. 40. 

14. To trisect any angle ABC^ 
From B, with any radius describe the arc AHC 
the angle ABC ; join A and C 
cutting the bisector in D ; with 
D as centre, and with DA as 
radius, describe the semicircle 
AGC, cutting the bisector in G ; 
and with the same radius, set 
off the points E and F from A 
and C ; join AG ; take the 
length AG and set it off from 
H along the line GB, which 
will give the point I ; join EI 
and FI, which will give the 
points J and K on the arc AHC ; 
join J and K with B, which wil 

1 This is one of the impossibiUties of geometry ; but this problem, devised 
by the author, gives an approximation so near, that the difference is imper- 
ceptible in ordmary geometrical drawing. 



bisect 




Fig. 41. 

trisect the angle ABC. 



GEOMETRICAL DRAWING AND DESIGN. 



Triangrles. 




D 



15. To construct an equilateral 
triangle on a given line AB. 

From A and B as centres, and 
with AB as radius, describe arcs 
cutting each other at C ; join C 
with A and B. Then ABC will 
be an equilateral triangle. 



16. On a given base AB to con- 
struct an isosceles triangle, 
the angle at vertex to be equal 
to given angle C. 

Produce the base AB to E, 
and at A construct an angle FAE 
making with AE an angle equal 
to C. Bisect the angle FAB 
by the line AD. From B draw 
a line making with AB an angle 
equal to DAB, and meeting 
AD in D. ADB will be the 
isosceles triangle required. 



17. On a given base AB, to con- 
struct an isosceles triangle, 
its altitude to be equal to a 
given line CD. 

Bisect AB at E, and erect 
a perpendicular EF equal in 
height to the given line CD ; 
join AF and BF, then AFB 
will be the isosceles triangle 
required. 



Fig. 44. 



PROBLEMS. 



21 



18. To construct a triangle, the three sides A, B, and C being given. 



Make the base DE equal to 
the given line A. From D as 
centre, and with radius equal to 
line B, describe an arc at F, 
and from E as centre, and with 
radius equal to line C, draw 
another arc, cutting the other at 
F ; join FD and FE, which will 
,give the triangle required. 



19. To construct a triangle with 
two sides equal to given lines 
A and B, and the included 
angle equal to C. 

Make an angle DEF equal 
to given angle C, in required 
position. Mark off EF equal 
to line A, and ED equal to 
line B ; join DF. DEF is the 
triangle required. 



20. To construct a triangle with 
a perpendicular height equal 
to AB, and the two sides 
forming the vertex equal to 
the given lines C and D. 

Through B draw the Hne EF 
at right angles to AB. From A 
as centre, and with radii equal 
to the lengths of the lines C 
and D respectively, draw arcs 
cutting the line EF ; join AE 
and AF, which will give the 
triangle required. 




Fig. 




Fig- 47- 



GEOMETRICAL DRAWING AND DESIGN. 




A 


/ 






Fig. 48. 






^T 






Fig. 50. 



21. To construct a triangle, on 
the given base AB, with 
one base angle equal to C, 
and the difference of the 
sides equal to given line D. 

At end A of the base, con- 
struct an angle equal to the 
given angle C. Cut off AF 
equal to line D, the given' 
difference of the sides ; join 
FB. Bisect FB at right 
angles by a line meeting AF 
produced in E ; join EB. 
AEB is the required triangle. 

22. To construct a triangle on 
a base equal to given line 
A, with vertical angle equal 
to D, and sum of the two rc- 

^ maining sides equal to BE. 

Draw the Hne BE, and at 
E construct an angle making 
with BE an angle equal to 
half the given angle D. 
From point B, with radius 
equal to given line A, draw 
an arc cutting EC at C ; join 
BC. Bisect CE at right 
angles by line FG ; join FC. 

BCF will then be the tri- 
angle required. 

23. To construct a triangle 
with two sides equal to 
the given lines A and B 
respectively, and the in- 
cluded median equal to 
given line C. 

Draw a triangle with the 
side DE equal to given line 
A, the side DF equal to 



PROBLEMS. 



2^ 



given line B, and the third side FE equal to twice the given 
median C. Bisect the line FE at G. Join DG and produce i*^ 
to H, making GH equal to DG ; join EH. 
DEH will be the triangle required. 



24. On a given base AB to 
describe a triangle similar 
to a given triangle DEF. 

Make angles at A and B 
equal respectively to the 
angles at D and E. Produce 
the lines to meet at C. Then 



ABC will 
required. 



25. 




be the triangle 



Quadrilaterals. 

To construct a square on a given base AB. 
C 



At point A erect AC perpen- 
dicular to AB and equal to it. 
With B and C as centres, and 
radius equal to AB, draw inter- 
secting arcs meeting at D ; join 
DC and DB. 

CABD is the square required. 



Fig. 52. 
26. To construct a square on a given diagonal AB. 

c 




Bisect AB at right angles by 
the line CD. Mark off EC and 
ED equal to EA and EB ; join 
CA and AD, and CB and BD. 

CADB is the square required. 




\-\E\ 



kl 



l^ig- 53 



24 



GEOMETRICAL DRAWING AND DESIGN. 



27. To construct a rectangle with sides equal to given lines A and B. 

Draw the line CD equal 
to given line A. At D erect 
a perpendicular DF equal to 
given line B. With C as 
centre and radius equal to 
line B, and with F as centre 
and radius equal to line A, 
draw arcs intersecting each 
other at E ; join EC and 
EF, which will give the rect- 
angle required. 

Fig. 54- 

28. To construct a rectangle with diagonal equal to given line A, 




F^'"~~ 




^11-'' -G 



and one side equal to 
given line B. 

Draw the line CD equal 
to given line A. Bisect CD 
at E. With E as centre, and 
with radius EC, describe a 
circle ; from C and D as 
centres, and with given line 
B as radius, set off the points 
F and G ; join CO, CD, 
DF, and FC. 

~ * CGDF is the required 

-~ rectangle. 

Fig. 55- 

29. To construct a rhombus with sides equal to given line A, and angle 

equal to given angle C. 

Make the base DE equal 
to given line A. At D con- 
struct an angle equal to given 
angle C. Set off DF equal 
to DE ; with F and E as 
centres, and radius equal to 
DE, draw arcs intersecting 
at G ; join FG and EG. 

DEGF will be the required 
rhombus. 



B— 




Fig. 56. 



PROBLEMS. 



25 



Bisect the angle BEF —-'" 



30. To draw a line bisecting the angle between two given 

converging lines AB and CD, when the angular point is 

inaccessible. /\ 

From any point E in AB, 
draw a line EF parallel to 
CD. 

by the line EG. At any 
point H between E and B, 
draw HL parallel to EG. 
Bisect EG and HL in M and 
N. Join MN, which, pro- 
duced, is the bisecting line C 
required. 
31. Through the given point A, to draw a line which would, if 

produced, meet at the same point as the given lines BC and 

DE produced. 




l-'ig- 57- 



Draw any convenient line 
FG; join A^FandAC. Draw 
any line HK parallel to FG. 
At H draw the line HL 
parallel to FA, and at K 
draw the line KL parallel to 
GA, cutting each other at L. 
Draw a line through L and 
A ; AU is the convergent 
line required. 

Circles. 

32. To find the centre of a 
circle. 

Draw any chord AB, and 
bisect it by a perpendicular 
DE, which will be a diameter 
of the circle. Bisect DE in 
C, which will be the centre 
of the circle. 




Fig- 59- 



26 



GEOMETRICAL DRAWING AND DESIGN. 



33. 




To draw a circle through three given points A, B, C. 

Join AB and BC. Bisect 
AB and BC by perpendi- 
culars cutting each other at 
D, which will be the centre 
of the circle. From D as 
centre, and DA as radius, 
describe a circle, which will 
then pass through the given 
points A, B, C, as required. 

Note.— The two following 
problems are constructed in 
the same manner. 

Fig. 60. 

34. To draw the arc of a circle throug-h three given points A, B, C. 

35. To find the centre of a circle from a given arc AC. 

36. At the given equidistant poihtS A, B, C, D; etc.. on a given arc, 
to draw a numher of radial lines, the centre of the circle being 
inaccessible. 

With the points A, B, C, 

D, etc., as centres, with radii 
larger than a division, des- 
cribe arcs cutting each other 
at E, F, etc. Thus, from A 
and C as centres, describe 

arcs cutting each other at E, 

Fig- 61. J T^ .-u r 

and so on. Draw the Imes 

BE, CF, etc., which will be the radial lines required. 

37. To draw the arc of a 

circle through three 

given points A, B, C, 

the centre of the circle 

being inaccessible. 

With A and C as 
centres, and with a radius 
equal to AC, draw inde- 
finite arcs. From the 
points C and A draw lines 
through B till they meet 





Fig. 62. 



PROBLEMS. ?.; 



the arcs in D and E. From D and E set off short equal 
distances on the arcs above and below ; join the divisions 
on arc AD to point C, and the divisions on arc CE to 
point A. Where the lines from corresponding points intersect, 
we obtain a point in the arc ; for instance, where the line from 
Ei intersects the line from Di, we get the point F ; and so on 
with the other points, by joining which with a fair curve, we get 
the required arc. 



EXERCISES. 

' 1. Draw two parallel lines 2|" long and if" apart. 

• 2. Draw a line 3.75" long; at the right-hand end erect a perpen- 
dicular 2.25" high ; then, 1.50" from it, another perpendicular i|" high ; 
and bisect the remaining length by a line 3" long, at right angles to it. 
^ 3. Draw a line 3|" long, and divide it into seven equal parts. 

4. Draw a line 2f " long ; from the left-hand end mark off a distance 
■equal to i^", and from the right-hand end a distance of g" ; draw 
another line 1.75" long, and divide it in the same proportion. 

5. Mark the position of three points A, B, and C — A to be if" 
from B, B to be 2^" from C, and C i|" from A ; and join them. 

6. Draw an angle equal to the angle ACB in the preceding question, 
and bisect it. 

7. Draw a right angle, and trisect it ; on the same figure construct 
and mark the following angles, viz. 15°, 30°, 45°, 60", 75°; and also 
74°, 22\\ and y]\\ 

8. Construct a triangle with a base l|" long ; one angle at the base 
to be 60°, and the side opposite this angle 2" long. 

9. Construct a triangle with a base 2.25" long, and altitude o't 
I-75"- 

10. On a base i|" long, construct an isosceles triangle ; the angle at 
its vertex to be 30°. 

11. Draw a scalene triangle on a base 2" long ; and construct a 
similar triangle on a base 1.75" long. 

12. On a base 2^" long draw a triangle with the angle at its vertex 
90°. 

13. Let a line 2.25" long represent the diagonal of a rectangle ; 
complete the figure, making its shorter sides |" long. 

14. Construct a rhombus with sides if" long, and one of its angles 
60°. 

15. Draw any two converging lines, and through any point between 
them draw another line which, if produced, would meet in the same 
point as the other two lines produced. 



28 



GEOMETRICAL DRAWING AND DESIGN. 



16. Fix the position of any three points not in the same line, and 
draw an arc of a circle through them. 

17. Draw an arc of a circle, and on it mark the position of any three 
points ; from these points, without using the centre, draw lines which, 
if produced, would meet in the centre of the circle containing the arc. 

18. Construct a triangle, the perimeter to be 5.6", and its sides in 
the proportion of 5 : 4 : 3. 

19. Draw two lines AB, AC containing an angle of 75''. Find a 
point P, f" from A 13 and ^" from AC. Complete the isosceles triangle, 
of which BAC is the vertical angle, and the base passes through P. 

(June, '00.) 

20. Draw the figure shown (Fig. 63) according to the given figured 
dimensions. (May, '97.) 




m 

\VV7 





Fig. 64. 




21. Draw the pattern (Fig. 64) according to the given dimensions. 

(April, '96.) 

22. Copy the diagram (Fig. 65) enlarging it to the dimensions 
figured. (April, '98.) 

23. Draw the figure shown in the diagram, (Fig. 66) making the side 
of the outer hexagon i^" long. (June, '98.) 




''^Qll 



E 



m 




Fig. 66. 



Fig. 67. 



Fig. 68. 



24. Draw the corner ornament shown (Fig. 67), using the figured 
dimensions, (April, '99.) 

25. Draw the given diagram of window tracery (Fig. 68) using the 
figured dimensions. The arch is ' equilateral'. (June, '99.) 



PROBLEMS. 



29 



26. Draw the given outline of window tracery (Fig. 69) using the 
figured dimensions. The arch is " equilateral," and all the arcs are of 
equal radius. (April, '00.) 



/XX ; 


11 


n^r 


C 




A 


A 


A 


"^ 




-f*-. 


>fs- 






1 1 


_J L_ 






nX,' 




Fig. 69. 




Fig. 70. 





27. Draw the given frame (Fig. 70), using the figured dimensions. 
The border is J" wide throughout. (April, '00.) 

28. Draw the given figure (Fig. 71), using the figured dimensions. 
(This problem is intended as an exercise in the use of T and set squares.) 

(June, '00.) 



- z' - ■*■ 

Fig. 71. 


• ' - J*' -.» 

Fig. 72. 



29. Draw the given figure (Fig. 72), using the figured dimensions. 
The spaces are ^" wide throughout. (June '99. ) 

30. Construct an isosceles right-angled triangle having its hypotenuse 
(or side opposite the right-angle) 2|" long, ^^'ithin it inscribe a square 
having one of its sides in the hypotenuse of the triangle. Measure and 
state, as accurately as you can, the length of one side of the square. 

(June, '01.) 
C. 



Q. 32. 



31. The given figure (Q. 31) is made up of a rectangle and semi- 
circles. Make a copy of it, using the figured dimensions. (June, '02.) 

32. Make a figure similar to the given figure (Q. 32) and having the 
height CD increased to 2f ". The centres of the arcs are given. 

(June, '03.) 





CHAPTER III. 



POLYGONS. 



Regular Polygons are figures that have equal sides and equal 
angles. To construct a regular polygon, we must have the length 

of one side and the number of 
sides ; if it is to be inscribed in 
a circle, the number of its sides 
will determine their length. 

If we take any polygon, regu- 
lar or irregular, and produce all 
its sides in one direction only, 
Fig. 73, we shall find that the 
total of all the exterior angles, 
shown by the dotted curves, is 
equal to 360°, or four right 
angles ; and if we join each 
angle of the polygon to any 
point in its centre, the sum of 
the angles at this point will also 
be 360°, and there will be as 
many angles formed in the centre 
as there are exterior angles. 
In regular polygons these angles at the centre will, of course, 
be equal to each other ; and if we produce the sides in one 
direction, as in Fig. 73, the exterior angles will be equal to each 
other; and as the number of angles at the centre is equal to the 
number of exterior angles, and the sum of the angles in each 
instance is equal, the angle at the centre must equal the exterior 
angle. 




POLYGONS. 



iiven line AB, for 



To construct any regular polygon on the 
example, a nonagon, 

360° -r 9 =40°. 
So if we draw a line at B, 
making an angle of 40° with 
AB produced, it will give us 

the exterior angle of the nona- r 

gon, from which it will be 
easy to complete the polygon. 

The perimeter of a polygon is sometimes given, e.g. Construct 
an octagon the perimeter of which is 6 inches. 



^.-i 



B 

• 74- 



inches. 



inches, inches. 

- 75 = I- 




Draw the line AB this length. It has been shown that the 
exterior angles and those at the centre of a regular polygon 
are equal, 360° -^ 8 = 45°. Pro- 
duce the Hne AB, and con- 
struct an angle of 45°. Make 
BC = AB. We now have three 
points from which we can 
draw the circle containing the 
required polygon (Prob. 33). 

The polygon could also be 
drawn, after finding the length of AB, by any of the methods 
shown for constructing a polygon on a given straight line. 

The centre of any regular polygon is the centre of the circle 
that circumscribes it. 

Any regular polygon can be inscribed in a circle by 
making angles at the centre equal to the exterior angle as 
above. 

If tangents to a circle circumscribing a regular polygon be 
drawn parallel to the sides of the inscribed polygon, or if 
tangents be drawn at the angles of the inscribed polygon, a 
similar figure will be described about the circle, and the 
circle will also be " described by," i.e. contained in, a similar 
figure. 

In the following problems two general methods are given for 
constructing regular polygons on a given line, and two for 



32 



GEOMETRICAL DRAWING AND DESIGN. 



inscribing them in a given circle ; but as these general methods 
require either a line or an arc to be first divided into equal 
divisions, the special methods for individual polygons are pre- 
ferable, which are also given. 

See also how to construct any angle without a protractor 
(Prob. 134), and its application to polygons. 



38. To inscribe in a circle, a triangle, square, pentagon, hexagon, 
octagon, decagon, or duodecagon. 

Describe a circle, and draw the two diameters AEand BD at 
right angles to each other ; join BA. AB is a side of a square. 

Set off on circumference, AF 
equal to AC ; AF is a side of a 
hexagon ; join AF and EF ; EF 
is a side of an equilateral 
triangle. With D as centre, 
and radius equal to EF, mark 
off G on EA produced. With 
G as centre, and radius equal to 
AC, set off H on circumference; 
join AH ; AH is a side of an 
octagon. With D and E as 
centres, and radius equal to 
DC, set off the points I and J on 
circumference ; join ID ; ID is 
a side of a duodecagon. With 
CG as radius, and J as centre, mark off K on diameter BD. 
With CK as radius, mark off on circumference from B the points 
L and M ; join BL and BM ; then BL is the side of a decagon, 
and BM is a side of a. pentagon. 




Approximate Constructions, 39 42. 

39. To inscribe any regular polygon in a given circle; 
for example, a heptagon. 

Describe a circle, and draw the diameter AB. Divide AB 



POLYGONS. 



33 



into as many equal divisions as there are sides to the polygon 
(in this instance seven). With 
A and B as centres, and with 
radius equal to AB, describe 
arcs intersecting at C. From C 
draw the line CD, passing through * 
the second division from A,^ till 
it meets the circumference at D. 
Join AD, which will give one side 
of the polygon ; to complete it, 
mark off AD round the circum- 
ference. 

This method of constructing 
polygons is due to the Chevalier 
Antoine de Ville (1628), and 
although useful for practical purposes, is not mathematically 
correct. 




Fig. 77. 



40. To inscribe any regular polygon in a given circle 
(second method) ; for example, a nonagon. 

Describe a circle, and draw the radius CA. At A draw a 
tangent to the circle. With 
A as centre, and with any 
radius, draw a semicircle, and 
divide it into as many equal 
parts as there are sides to 
the polygon (in this instance 
nine). 

From point A draw lines 
through each of these divisions 
till they meet the circum- 
ference. Join these points, 
which will give the polygon 
required. 

1 Whatever number of sides the polygon may have, the Hne CD is ahvays 
drawn through the second division from A. 

C 




34 



GEOMETRICAL DRAWING AND DESIGN. 



41. On a given line AB, to describe a regular polygon; 
for example, a heptagon. 

Produce AB to C. With A as centre, and AB as radius, 

describe a semicircle, and 
divide it into as many equal 
divisions as there are sides to 
the polygon (in this instance 
seven). Join A with the second 
division from C in the semi- 
circle,^ which will give point 
D. Join AD. Through the 
points DAB describe a circle. 
Set off the distance AD 
round the circumference, and 
join the points marked, which will give the required polygon. 




42. On a given line AB, to describe a regular polygon 

(second method) ; for example, a pentagon. 

At point B erect a perpendicular equal to AB. With B as 

centre, and radius BA, draw the 
quadrant AC, and divide it into 
as many equal divisions as 
there are sides to the polygon 
(in this case five). 

Join the point B with the 

second division from C.^ Bisect 

/ AB at D, and erect a perpendi- 

/ cular till it meets the hne drawn 

from B to the second division, 

which will give point E. From 

E as centre, and with radius 

EB, describe a circle. From 

AB, mark off round the circum- 




B 



Fig. 8o. 

point A, with distance 



1 The line AD is always drawn to the second division from C, whatever 
number of sides the polygon may contain. 



POLYGONS. 



35 



ference the points of the polygon, 
will give the pentagon required. 



Join these points, which 



43. On a given straight line AB, to construct a regular 
pentagon. (True construction.) 

At B erect BC perpen- 
dicular to AB, and equal 
to it. Bisect AB in D. 
With D as centre, and DC 
as radius, draw the arc CE 
meeting AB produced in E. 
With A and B as centres, 
and radius equal to AE, 
draw arcs intersecting at 
F. With A, B, and F as 
centres, and radius equal 
to AB, draw arcs intersect- 
ing at H and K. Join AH, 
HF, FK, and KB. Then 
required. 




AHFKB will be the pentagon 



44. On a given straight line AB, to construct a regular hexagon. 



With A and B as centres, 
and radius AB, draw the arcs 
intersecting each other at C. 
With C as centre, and with 
the same radius, draw a circle. 
With the same radius com- 
mencing at A, set off round 
the circle the points D, E, F, 
G. Join AD, DE, EF, FG, 
and GB, which will give the 
hexagon required. 




E.-'---. f 




Fig. 82. 



36 



GEOMETRICAL DRAWING AND DESIGN. 



45. In a given circle to inscribe a regular heptagon 
(approximately). 




Draw any radius AB, and 
bisect it in C. Through C 
draw DE perpendicular to 
AB. With CD as radius, 
commencing at E, set off 
round the circle the points 
F, G, H, K, L, and M, by 
joining which we get the 
required heptagon. 



Fig. 83. 



46. On a given line AB, to construct a regular heptagon 
(approximately). 

With B as centre, and BA as radius, draw a semicircle 

meeting AB produced in 

C. With centre A, and 
radius AB, draw an arc 
cutting the semicircle in 

D. Draw DE perpendi- 
cular to AB. With C as 
centre, and radius equal 
to DE, draw an arc cut- 
ting the semicircle in F. 
Join BF. From the three 
points A, B, F, find the 
centre of the circle H 
(Prob. 33). With H as 
centre, and radius HA, 




Fig. 84. 



draw a circle. With AB as radius commencing at F, set off 
on the circle the points K, L, M, and N, by joining which we 
get the heptagon required. 



POLYGONS. 



37 



47. On a given line AB, to construct a regular octagon. 

At A and B erect the perpendiculars AC and BD. Produce 
AB to E. Bisect the 
angle DBE by the line 
BF. Make BF equal to 
AB. From F draw the 
line FH parallel to AB, 
and make KH equal to 
LF. Join AH. Set off 
on AC and BD, from 
points A and B, a length 
equal to H F, which will 
give the points C and D. 
With the points C, D, H, 
and F as centres, and a 

radius equal to AB, draw ^,- g. 

arcs intersecting at M 

and N. Join HM, MC, CD, DN, and NF, which will give the 
octagon required. 




48. In a given circle to inscribe a nonagon. 

Draw the diameters AB and CD perpendicular to each other. 
With A as centre, 
and radius AE, draw 
the arc cutting the 
circle in F. With B 
as centre, and radius 
BF, draw the arc 
cutting CD produced 
in G. With G as 
centre, and radius 
GA, draw the arc 
cutting CD in H. 
With HC as radius, 
commencing at A, 
set off on the circle 
the points K, L, M, 
N, O, P, Q, and R, by joining which we get the nonagon required. 




-C 



3^ 



GEOMETRICAL DRAWING AND DESIGN. 



49. In a given circle to inscribe a regular undecagon. 

Draw the two diameters AB and CD cutting each other in E. 

With A as centre, and AE 
as radius, draw an arc cut- 
ting the circle in H. With 
D as centre, and DE as 
radius, draw an arc cutting 
the circle in F. With F as 
centre, and FH as radius, 
draw the arc cutting CD in 
K. Join HK. Then HK 
is equal to one side of the 
undecagon. With HK as 
radius, starting from H, set 
off en the circle the points of 
the required undecagon, and join them. 




EXERCISES. 

1. Inscribed in circles of varying diameters, draw the regular 
polygons, from a pentagon to a duodecagon, by a general method and 
figure the angles formed by their sides. 

2. In circles of various radii, draw all the preceding polygons by 
special methods. Join their angles to the centre of the circle by radii, 
and figure the angles between the radii. 

3 On lines varying in length, draw the same polygons by a general 
method. 

4. Construct on lines of different lengths the same polygons by 
special methods. 

5. Construct an irregular hexagon from the following data : Sides, 
AB I", BC if, CD f", DE ih", EF i|" ; Angles, ABC 140°, BCD 
130°, CDE 110°, DEF 120°. 

6. Construct an irregular pentagon from the following data : Sides, 
AB 1.25", BC 1.3", CD 1.2", DE 1.3", EA 1.4"; Diagonals, AC 
1.8" AD 1.6". 

7. Construct a regular polygon with one side i" in length and one 
angle 140°. 

8. How many degrees are there in each of the angles at the centre 
of a nonagon ? 

9. Construct a regular polygon on the chord of an arc of 72°. 



POLYGONS. 



39 




10. Inscribe in any given circle an irregular heptagon whose angles 
at the centre are respectively 52°, "j^^^ 45°. 63°, 22°, 36'', and 69°. 

11. Construct a regu]ar octagon of i" side, and a second octagon 
having its angles at the middle points of the sides 
of the first. (May, '97.) 

12. Draw a figure similar to the one shown 
(Fig. 88), the points of the star being at the angles 
of a regular heptagon inscribed within a circle of 
1 1" radius. (June, '97.) 

'3. Within a circle of i^" radius inscribe a 
regular nonagon. Within tne nonagon inscribe a 
rectangle having all its angles in the sides of the 
nonagon, and one of its sides i^" long. 

14. Construct a regular pentagon of iV side. Describe five circles 
off" radius, having their centres at the five angles of the pentagon. 

(June, '98.) 

15. Draw the given figure (Q. 15), making the radius of the outer 
circle i\". (April, '01.) 

16. Construct a regular pentagon of 2" side, and a similar pentagon 
of 2" diagonal. The two figures should have the same centre. 

(N.B. — The protractor may not be used for obtaining the angle of 
the pentagon.) (June, '02.) 



Fig. £8. 
(April, 98.) 




Q. 15- 





Q 19. 



17. Copy the diagram (Q. 17), using the figured dianensions. 

(June, '02.) 

18. Within a circle of I "5" radius inscribe a regular nonagon. With 
the same centre describe a regular nonagon of 075" side. (May, '03.) 

19. Copy the given figure (Q. 19). The straight lines are to form a 
regular pentagon of I 'S" side, and the five equal segments are to have 
their arcs tangential at the angles of the pentagon. Show all your 
constructions clearly. ' (June, '03.) 



CHAPTER IV. 



INSCRIBED AND DESCRIBED FIGURES. 
50, To inscribe an equilateral triangle in a given circle ABC. 



Find the centre E (Prob. 
H 




32), and draw the diameter DC. 
With D as centre, and DE as 
radius, mark off the points A 
and B on the circumference of 
the circle. Join AB, BC, and 
CA. Then ABC will be the 
inscribed equilateral triangle 
required. 



51. To describe an equilateral 
triangle about a given circle 

ABC. 
At the points A, B, and C 
draw tangents to the circle 



(Prob. 84), and produce them till they meet in the points F, G, and 
(^ H. Then the equilateral triangle 

FGH will be described about 
the circle ABC, as required. 




52. 



To inscribe a circle in a 
given triangle ABC. 



Bisect the angles CAB and 
ABC by lines meeting in D. 
From D let fall the line DE, 
perpendicular to AB. With D 
as centre, and DE as radius, 
inscribe the circle required. 



INSCRIBED AND DESCRIBED FIGURES. 



41 



53. To describe a circle about 
a given triangle ABC. 
Bisect the two sides AB 
and AC perpendicularly by 
lines meeting in D. With D 
as centre, and DA as radius, 
describe the required circle. 

54. To describe an equilateral 
triangle about a given 
square ABDC. 

With C and D as centres, 
and with CA as radius, de- 
scribe arcs cutting each other 
at E. With E as centre, and 
with the same radius, mark 
off the points F and G on 
these arcs. Join CF and DG, 
and produce them till they 
meet in the point H, and the 
base AB produced in K and 
L. Then HKL is the re- 
quired equilateral triangle. 

55. In a given triangle ABC, 
to inscribe an oblong hav- 
ing one of its sides equal 
to the given line D. 

From A, along the base 
AB, set off AE equal to the 
given line D. From E draw 
the line EF parallel to AC. 
Through F draw the line FG 
parallel to the base AB. 
From G and F draw the lines 
GH and FL perpendicular to 
AB. GFLH is the oblong 
required. 




Fig. 93- 



42 



GEOMETRICAL DRAWING AND DESIGN. 



To inscribe a square in a given circle. 

C y\ Draw any two diameters AB 

and CD at right angles to each 
other. Join the extremities of 
these diameters. ACBD is the 
inscribed square required. 

57. To describe a square about 

a given circle. 
At the points A, C, B, D draw 
tangents meeting each other 
at the points E, F, G, H. (Prob. 
84). EFHG is the described 
square required. 

To inscribe a circle in a given square. 

Draw the diagonals AB and 
CD intersecting each other at 
E. From E draw EF perpendi- 
cular to AD. With E as centre, 
and EF as radius, draw a 
circle. This will be the in- 
scribed circle required. 

59. To describe a circle about a 
given square. 
With centre E, and radius 
EA, draw a circle. This will be 
the described circle required. 





60. To inscribe a square in 
a given rhombus. 
Draw the two diagonals 
AB and CD. Bisect the 
angles AEC, AED, and 
^ B produce the lines each way 
till they meet the sides of 
the rhombus in the points 
F, G, H, L. Join FG, GL, 
LH,andHF. FGLH is the 
inscribed square required. 



INSCRIBED AND DESCRIBED FIGURES. 



43 



61. To inscribe a circle in a 

given rhombus. 
Draw the diagonals AB 
and CD intersecting each 
other in E. From E draw 
the line EF peipendicular to^ 
AD (Prob. 7). With E as 
centre, and EF as radius, 
draw a circle. This is the 
inscribed circle required. 

62. To inscribe an equilateral 
triangle in a given square 
ABDC. 

With B as centre, and radius 
BA, draw the quadrant AD ; 
and with same radius, with A 
and D as centres, set off on 
the quadrant the points F and 
E. Bisect AE and FD, and 
through the points of bisection ■^~-- 
draw lines GB and HB, cut- 
ting the given square in G and 
H. Join GH. BGH is the 
inscribed equilateral triangle 
required. 

63. To inscribe an isosceles 
triangle in a given square 
ABDC, having a base 
equal to the given line E. 

Draw the diagonal AD. 
From A, along AD, set off 
AF equal to half the given 
base E. Through F draw 
the line GH perpendicular to 
AD (Prob 5). Join GD and 
HD. GDH is the inscribed 
isosceles triangle required. 




Fig. 99. 



44 



GEOMETRICAL DRAWING AND DESIGN. 



64. To inscribe a square in a given trapezium ACBD which 
has its adjacent pairs of sides equal. 

Draw the two diagonals 
AB and CD. From point C 
set off CE perpendicular and 
equal to CD. Join EA by a 
line cutting CB in G. Draw 
the hne GF parallel to AB. 
From points F and G draw 
the lines FH and GK parallel 
to CD. JoinHK. FGKH is 
the inscribed square required. 

65. To inscribe a circle in a 

given trapezium ACBD 

which has its adjacent 

pairs of sides equal. 

Bisect any two adjacent 

angles, as ADB and DBC, 

by lines meeting in E. From 

E draw EF perpendicular to 

CB. With E as centre, and 

EF as radius, draw a circle. 

This will be the inscribed 

circle required. 

To insert a rhombus in a given rhomboid ABDC. 

>^ Draw the diagonals AD 

C ,'P P and CB intersecting at E. 

Bisect two adjacent angles, 
as CED and DEB, by lines 
cutting the given rhomboid 
in F and G, and produce 
these lines to K and H. 
Join FG, GK, KH, and 
HF. FGKH is the in- 
scribed rhombus required. 
67. To inscribe an octagon in a given square ABDC. 
Draw the two diagonals AD and CB intersecting each other 
in E. With A as centre, and AE as radius, mark off the points F 





INSCRIBED AND DESCRIBED FIGURES. 



45 



and G on the sides ot the 
square. Proceed in the same 
manner with the angles B, C, 
and D as centres, which will 
give the eight points required 
on the given square, by joining 
which we obtain the required 
inscribed octagon. 

68. To inscribe a square in a 
given hexagon ABCDEP. 

Draw the diagonal EB ; 
bisect it in G, and draw HK 
perpendicular to it. Bisect 
two adjacent angles, as BGK 
and EGK, by lines meeting 
the hexagon in O and M. 
Produce these two lines till 
they meet the opposite sides 
of the hexagon in L and N. 
Join LM, MO, ON, and NL. 
LM ON is the inscribed square 
required. 

69. To inscribe four equal 
circles in a given square 
ABDC ; each circle to touch 
two others, as well as two 
sides of the given square. 

Drav.' the two diagonals 
AD and CB intersecting at E. 
Bisect the sides of the square 
in the points F, G, H, and L 
(Prob.i). Draw FH and GL. 
Join FG, GH, HL, and LF, 
which will give the points M, 
O, P, and R. From M draw 
MN parallel to AB. With 
M, O, P, and R as centres, and 
radius equal to MN, describe 
the four inscribed circles re- 
quired. 




46 



GEOMETRICAL DRAWING AND DESIGN. 




70. To inscribe four equal circles in a given square ABDC ; 
each circle to touch two others, and one side only of the 
given square. 

Draw the two diagonals 
AD and CB intersecting in 
the point E. Bisect the sides 
of the square in the points 
F, G, H, and L. Join FH 
and GL. Bisect the angle 
GDE by a hne meeting GL 
in M. With centre E, and 
radius EM, mark off the 
points N, O, and P. With 
the points M, N, O, and P as 
centres, and radius equal to 
MG, describe the four in- 
scribed circles required. 

71. To inscribe three equal circles in a given equilateral triangle 
ABC ; each circle to touch the other two, as well as two 
sides of the given triangle. 

^ Bisect the two sides of the 

triangle at right angles by- 
lines meeting at the centre D. 
Draw a line from point C 
through the centre D till it 
meets the base at E. Bisect 
the angle BEC by a line 
meeting DB in F. With D 
as centre, and DF as radius, 
mark off the points K and 
L. From F draw the line 
FH perpendicular to AB. 
With the points F, K, and L 
as centres, and with a radius 
equal to FH, draw the three inscribed circles, required. 

72. In a given equilateral triangle ABC, to inscribe three equal 
circles touching each other and one side of the triangle only. 

Bisect two sides of the triangle at right angles by lines 




INSCRIBED AND DESCRIBED FIGURES. 



47 



meeting at the point D. Join CD and produce it to E, 

Bisect the angle DBE by a line cutting CE in F. With D as 

centre, and DF as radius, mark off the points G and H, 

From the points F, G, and 

H as centres, and with a .^ 

radius equal to FE draw 

the three inscribed circles 

required. 




73. In a given equilateral 
triangle ABC, to inscribe 
six equal circles touching 
each other. 

Having drawn the three 
circles according to the last 
problem, draw lines through 

G and H, parallel to the sides of the triangle, till they meet the 
lines bisecting the angles in the points L, M, and N. These 
points are the centres of the three circles which will complete 
the six inscribed circles required. 



74. In a given octagon to inscribe four equal circles touching 
each other. 

Draw the four diagonals 
meeting in C. Bisect the 
angle ABC by a line inter- 
secting AC in D. With C 
as centre, and CD as radius, 
mark off the points F, G, and 
H. From D draw the line 
DL perpendicular to KC. 
With D, F, G, and H as 
centres, and a radius equal 
to DL, draw the four in- 
scribed circles required. 

Fig. log 




48 



GEOMETRICAL DRAWING AND DESIGN. 



75. In a given circle to draw four equal circles touching 
each other. 

Find centre of circle E 
(Prob. 32). Draw the two 
diameters AB and CD at 
right angles to each other. 
At A and D draw tangents 
to the circle to meet at the 
) point F (Prob. 84). Join FE. 
Bisect the angle EFD by a 
line cutting CD in G. With 
E as centre, and EG as 
radius, mark off the points 
H, K, and L. With G, H, 
K, and L as centres, and 
with a radius equal to GD, 
draw the four inscribed circles required. 




76. In a given circle to inscribe any number of equal circles 
touching each other. For example, five. 

Find the centre C (Prob. 32). Divide the circumference into 
five equal parts (Prob. 38), and draw the five radii to meet the 

circumference in the points 
M, N, O, P, and R. Bisect 
the angle MCN by a line 
meeting the circumference in 
A. Through A draw a line 
at right angles to CE till it 
meets the Hnes CM and CN 
produced in B and D. Bisect 
the angle DBC by a Hne to 
meet CA in E. With C as 
centre, and CE as radius, 
draw a circle, and bisect 
each arc on this circle, 
between the five radii, in 
the points F, G, H, and L. 
With E, F, G, H, and L as centres, and a radius equal to EA, 
draw the five inscribed circles required. 




INSCRIBED AND DESCRIBED FIGURES. 



49 



77. About a given circle A, to describe six circles equal to it, 
touching each other as well as the given circle. 



Find centre of circle A (Prob. 
32). With A as centre, and a 
radius equal to the diameter of 
the given circle, draw the circle 
BCDEFG. Draw the diameter 
BE. Take the diameter of the 
giv^en circle and mark off from 
E the points D, C, B, G, and 
F. With each of these points 
as a centre, and a radius equal 
to that of the given circle, draw 
the six circles required. 




Fig. 112. 



EXERCISES. 

1. Within a given circle inscribe a square, and about the same 
circle describe an equilateral triangle. 

2. Construct a rhombus with sides i^" long, and its shorter 
diagonal 1.75" ; inscribe a circle within it, and let the circle circum- 
scribe an equilateral triangle. 

3. Construct a trapezium with two of its sides if" and two i|" 
respectively, and with its longer diagonal 2^" ; inscribe within it a 
square, and let the square circumscribe an equilateral triangle. 

4. Draw a regular hexagon with i" sides and let it circumscribe a 
square ; inscribe a regular octagon within the square. 

5. Draw any triangle, and describe a circle about it. 

6. Construct a square of 2^" sides, and in it inscribe an isosceles 
triangle with a if" base ; inscribe within the triangle a rectangle, one 
side of which is i|". 

7. Within a square of 1.75" sides, inscribe an isosceles triangle with 
angle at vertex 60° ; inscribe a circle within the triangle. 

8. Within a circle of any radius, inscribe a regular duodecagon, and 
let it circumscribe a hexagon. 

9. Construct a rhomboid with sides 2" and i^", its contained angle 
to be 60° ; inscribe within it a rhombus. 

10. Within an equilateral triangle of 3" sides, inscribe a circle, and 
within it 3 equal circles. 



50 GEOMETRICAL DRAWING AND DESIGN. 

11. Draw a circle of i|" radius ; inscribe within it an equilateral 
triangle ; inscribe within the triangle three equal circles touching each 
other and each one side of the triangle only. 

12. Construct a square of 2" sides, and let it circumscribe four equal 
circles ; each circle to touch two others, as well as two sides of the 
square. 

13. Construct a square with sides of 2.3", and inscribe four equal 
circles within it ; each circle to touch two others, as well as one side 
only of the square. 

14. Within a triangle of 2.7" sides, inscribe six equal circles. 

15. • Within a circle of 1.7" radius, inscribe seven equal circles. 

16. Draw two concentric circles, and between them, six equal circles, 
to touch each other as well as the two concentric circles. 

17. In a decagon, inscribe five equal circles. 

18. Construct an equilateral triangle of 2f" side. Bisect all three of 
its sides and join the points of bisection. Within each of the four 
equilateral triangles thus formed inscribe a circle. (April, '98.) 

19. Draw an equilateral triangle, a scalene triangle, a right-angled 
triangle, and an oblong ; a trapezium, and a regular polygon of 
eleven sides, each in a 2-inch circle, and write the names to each. 

(May, '96.) 

20. Draw each of the following figures in a separate 2-inch circle, an 
isosceles triangle, an obtuse-angled triangle, an acute-angled triangle, a 
square, a rhombus, and a rhomboid, and write the name to each. 

(May, '97.) 

21. Draw a square^ an ob.'ong, and a trapezium ; a pentagon^ a 
hexagon, an octagon, and t7U0 parallel straight lines, each in a 
separate 2-inch circle, and write the name to each. (April, '98.) 

22. About a square of \" side describe a triangle having one of its 
angles 60° and another 70°. (May, '97.) 

23. Within a rhombus, sides 2|", one angle 60°, inscribe an ellipse 
touching the sides of the rhombus at iheir middle points. (June, '97.) 

24. Construct a rectangle 2V' x if". Within it inscribe two other 
rectangles, each similar to the first, concentric with it and having their 
longer sides ij" and i" long respectively. (April, '98.) 

25. Within an equilateral triangle of 3" side inscribe three equal 
circles each touching the two others, and tivo sides of the triangle. 

(May, '97.) 

26. Construct a square of i|" side, and within it inscribe a rectangle 
having one of its angles on each side of the square and one of its sides 
i"long. (June, 97.) 

27. Construct a triangle, sides i^", 2", and 2^", and within it 



INSCRIBED AND DESCRIBED FIGURES. 51 

inscribe an equilateral triangle having its three angles in the three 
sides of the first triangle. (June, '98.) 

28. Within a square of if" side inscribe a regular octagon having all 
its angles in the sides of the square, (April, '99.) 

29. Construct a regular heptagon of i" side, and within it inscribe an 
equilateral triangle. (June, '99.) 

30. Construct a quadrilateral ABCD from the following data : — 

Sides— AB=ii", BC=i4" 
Angles— ABC =105°, BAD = 75° 

The four angles of the figure all lie in the circumference of a circle. 

(April, '98.) 

31. Within a circle of i|" radius inscribe a regular pentagon. 
About the same circle describe another regular pentagon, having its 
sides parallel to those of the inscribed pentagon. (April, '96.) 

32. Within a circle of 2|" diameter, inscribe four equal circles each 
touching the given circle and two of the others. (June, '98.) 

33. Within a circle of i|" radius inscribe a regular heptagon. Draw 
a second similar heptagon, of which the longest diagonals are 2" long. 

(April, '99.) 

34. Within a circle of i|" radius, inscribe a regular hexagon. 
Within the hexagon inscribe three equal circles touching each other, 
and each touching two sides of the hexagon. (June, '99.) 

35. Construct a rhombus having its sides 2" long and one of its 
angles 75°. Within it inscribe two equal circles touching each other, 
and each touching two sides of the rhombus. (April, '99. ) 

36. Within a circle of 2|" diam. inscribe a regular pentagon. Draw 
also a second regular pentagon concentric with the first one, its sides 
being parallel to those of the first and I5" long. (June, 1900.) 

37. The sides of a triangle are i", i^" and i^" long. About this 
triangle describe a circle, and about the circle describe a triangle of the 
same shape as the given one. The points of 
contact must be found and clearly shown. 

(April, 1900.) 

38. Describe a circle touching both the 
given circles (Fig. 113) and passing through 
the point P on the circumference of the 
larger one, three times the size of figure. 

(April, 1900.) 

39. About a circle of i" diam. describe Fig. 113. 
six equal circles, each touching the given 

circle and two of the others. Then describe a circle touching and 
enclosing all six of the outer ones. 




52 



GEOMETRICAL DRAWING AND DESIGN. 



40. Make an enlarged copy of the given diagram (Fig. 114), using 
the figured dimensions. (June, '98. ) 





Fig. 115. 

41. Copy the diagram (Fig. 115) according to the given dimensions. 
Show clearly how the centre of the small circle is determined. 

(June, '98.) 

42. Within a regular hexagon of ii"side, inscribe a square having 
all its angles in the sides of the hexagon. Within the square inscribe 
four equal circles, each touching two of the others and two sides of the 
square. (April, '01.) 

43. About a circle of 075" radius describe an equilateral triangle. 
Describe three equal circles touching the given circle and having their, 
centres at the angles of the triangle. Determine the points of contact. 

(April, '02.) 

44. Draw the given figure (Q. 44), making the side of the square 2^" 
long. Show clearly how all the points of contact are determined. 

(April, '02.) 




Q. 44. 



46. About a circle or o*8" radius describe a rhombus having one oi 
its angles 54°. Each side of the rhombus touches the circle. Deter- 
mine the four points of contact. (June, '02. ) 

46. Describe a semicircle of I '5" radius. Within it inscribe two 
circles, each touching the other, and also touching the circumference 
and diameter of the semicircle. (June, '03.) 



CHAPTER V. 

FOILED FIGURES. 

Foiled figures are constructed on the regular polygons, and 
are of two kinds : viz. those having tangential arcs, and those 
having adjacent diameters. 

Problem 78 is an illustration of the former kind. The arcs 
simply touch, and if produced would not intersect each other. 
The angles of the polygons are the centres of the circles con- 
taining the arcs. 

Problem 79 is an illustration of the latter kind. The sides of 
th -polygon form the diameters of the semicircles, the centre 

each side being the centre of the circle containing the 
arc. If these slyc^ were produced, they would intersect each 
other. 



78. To construct a foiled figure 
about any regular polygon, 
having tangential arcs. For 
example, a hexagon. 

The Hexafoil. — Bisect one side 
AB in C. With each of the 
angular points as centres, and 
AC as radius, draw the six arcs, 
as required. 




Fig. 116. 



54 



GEOMETRICAL DRAWING AND DESIGN. 



79. To construct a foiled figure about any regular polygon, having 
adjacent diameters. For example, a pentagon. 



Tlie Cinquefoil. — Bisect 
each side of the pentagon 
in the points C, D, E, F, and 
G. With each of these points 
as centres, and with a radius 
equal to CA, draw the five 
arcs required. 

Note. — If these arcs are 
to have a stated radius, the 
length of the hne AB, in each 
instance, will be twice the 
required radius. 




In a given equilateral triangle ABC, to inscribe a trefoil. 

The Trefoil. — Bisect the 
angles CAB and ABC by 
lines produced till they meet 
in L, and the sides of the 
triangle in D and E. From 
C, through centre L, draw the 
line CF. Join DE, EF, and 
FD. From G draw GI per- 
pendicular to AC. WithG,H, 
and K as centres, and a radius 
equal to GI, draw the three 
arcs till they meet each other, 
which will form the trefoil 
required. 




81. Within a given circle, to inscribe three equal semicircles having 
adjacent diameters. 

Find centre of circle A (Prob. 32). Draw the diameter BC, and 



FOILED FIGURES. 



55 



the radius AD perpendicular to it (Prob. 5). Trisect the angle 
BAD in E and F (Prob. 13). 



Set off DG equal to DF. 
Join FA and GA by lines 
produced to L and H. Join 
EG by a line cutting FL in 
M. With A as centre, and 
AM as radius, set off the 
points N and O. Join MN, 
NO, and OM, which are the 
diameters of the required 
semicircles. With R, P, and 
S as centres, and a radius 
equal to RM, draw the three 
semicircles required. 




82. In a given square ABDC, to inscribe four semicircles having 
adjacent diameters. 

The Quatrefoil. — Draw the 
diagonals AD and CB. 
Bisect each side of the square 
in the points E, F, G, and 
H, and join EF and GH. 
Bisect HD in K and FB in 
L, and join KL, cutting GH 
in N. With P as centre, and 
PN as radius, mark off the 
points M, O, and R, and 
join NM, MO, OR, and RN, 
which will form the diameters 
of the required semicircles. 
With S, T, U, and V as 
centres, and with a radius 
equal to SN, draw the four 
semicircles required. 




56 



GEOMETRICAL DRAWING AND DESIGN. 



83. Within a given circle to inscribe any number of equal semi- 
circles having adjacent diameters. For example, seven. 

The Heptafoil. — Find centre C (Prob. 32). Divide the circle 

into as many equal parts 
as the number of semi- 
circles required (in this 
case seven), and from 
each of these points, A, 
B, D, E, F, G, and H, 
draw diameters. From 
the point A draw the 
tangent AM (Prob. 84), 
and bisect the angle CAM 
by a Hne cutting CK in 
N (Prob. 12). With C as 
centre, and CN as radius, 
mark off the points O, P, 
R, S, T, and U. Join each 
of these points to form the polygon. From the centre of each 
side of the polygon, with a radius equal to half of one of its 
sides, draw the seven semicircles required. 




EXERCISES. 

1. Construct an equilateral triangle of i^" sides, and about it describe 
a trefoil having tangential arcs. 

2. Construct a pentagon of f" sides, and about it describe a cinque- 
foil having adjacent diameters. 

3. Draw a pentagon of ^" sides, and about it construct a cinquefoil 
having tangential arcs. 

4. In a circle of i^" diameter, draw nine equal semicircles having 
adjacent diameters. 

5. In a circle of 1" diameter, inscribe a quatrefoil having tangential 
arcs. 

6. Construct a regular decagon in a circle of |" radius, and within it 
inscribe a cinquefoil. 

Note. — Foiled figures can be inscribed in all the regular polygons 
that have an even number of sides, by first dividing them into trapezia 
and then proceeding by the method shown in Prob. 65 ; or they can be 
drawn in circles divided into any number of equal sectors, by Prob. 106. 



CHAPTER VI. 
TANGENTS AND TANGENTIAL ARCS 



84. To draw a tangent to a 
given circle at a given 
point A. 

Find the centre of the" 
circle C (Prob. 32). Join AC. 
From A draw the hne AB 
perpendicular to AC, and 
produce it. Then AB is the 
tangent required. 




Fig. 122 



85. To draw a tangent to a 
given circle from a given 
point A outside it. 

Find the centre C (Prob. 
32). Join AC, and bisect 
in B. From B as centre, 
and with BA as radius, de- 
scribe" a semicircle cutting 
the circumference in D. Join 
AD, and produce it. Then 
the line AD is the tangent 
required. 




Fig. 123. 



58 



GEOMETRICAL DRAWING AND DESIGN. 





86. To draw a tangent to the 
arc of a circle at a given 
point A without using the 
centre. 

Draw the chord AB and 
bisect it in C). At C erect a 
perpendicular to AB cutting 
the arc in D. Join AD. 
Make the angle DAE equal 
to DAC. Then EA produced 
is the tangent required. 

87. To draw a circle with 
radius equal to line D to 
touch two straight lines 
forming a given angle 
ABC. 

Bisect the angle ABC by 
the line BE. Draw the hne 
FIT parallel to BA (Prob. 4), 
and at a distance equal to 
given line D from it. Where 
FH intersects BE will be the 
centre of the circle. With 
E as centre, and D as radius, 
draw the circle required. 



88. To draw tangents to a 

circle from a given point 

A, the centre of the circle 

not being known. 

From point A draw any 

three secants to the circle. 

as CB, DE, and GF. Join 

BE and DC, DG and FE, 

by lines intersecting in the 

points H and K. Draw a 

line through H and K till 



TANGENTS AND TANGENTIAL ARCS. 



59 



it meets the circumference in L and M. Join AL and AM, 
which will be the tangents required. 



89. In a given angle CAB, to inscribe a circle wliich shall pass 
through a given point D, 

Bisect the angle CAB, 
by the line AE (Prob. 62). 
Take any convenient 
point F in AE, and from 
F draw the line FG per- 
pendicular to AC (Prob. 
7). With F as centre, 
and FG as radius, draw a 
circle. Join DA, cutting 
the circle in H. From 
the given point D draw 
DK parallel to HF. With 
K as centre, and KD as 
radius, draw the required 
circle. 




Fig. 127. 



90. To draw a circle which shall pass through the given point 
A and touch a given line BC in D. 



At the given point D 
erect the Hne DE perpen- 
dicular to BC (Prob. 5), 
and join AD. At point A 
construct an angle DAE 
equal to the angle ADE 
(Prob. 11). AE intersects 
DE in E. With E as 
centre, and ED as radius, 
draw the required circle. 




6o 



GEOMETRICAL DRAWING AND DESIGN. 



91. To dra-w a circle which shall pass through the two given 
points A and B and touch the given line CD. 

Join the two given 
points AB, and produce 
the line till it meets the 
given line CD produced 
in E. On AE describe 
the semicircle EFA ; at B 
draw BF perpendicular to 
AE. From E along the 
line ED, set off EG equal, 
to EF. Then G, B, A 
are three points in thd 
required circle, which can 
be drawn as required 
(Prob. 33). 




92. To draw four equal circles, with radius equal to given line 
E, to touch two given lines AB and CD, which are not parallel. 




Bisect the two adjacent, 
angles by the lines FGj 
and HK. Draw the lines 
LM and NO parallel tc 
the given line CD, at a 
distance from it equal tc 
the given radius E (Prob. 
4). Where these lines 
intersect the bisectors, we 
get the points S, R, T, and 
P. With the points R, S, 
T, and P as centres, and 
with a radius equal to E, 
draw the four circles re- 
quired. 



TANGENTS AND TANGENTIAL ARCS. 



6i 



93. To draw an inscribed and an escribed circle, tangential to 
three given straight lines, forming a triangle ABC. 

Note. — An escribed circle is also called an excircle. 



Bisect the angles BAG and ACB 
by lines intersecting in F. From 
F draw the line FH perpendicular 
to AD. With F as centre, and 
FH as radius, draw the inscribed 
circle required. Bisect the angle 
BCD by a line cutting the line 
AG in E. Draw the line EK per- 
pendicular to AL. With E as 
centre, and EK as radius, draw the 
escribed circle required. 




94. A principle of inscribed and escribed circles. 

If a triangle ABC be taken, and 
AF, BD, and CE be lines drawn 
from the three angles perpendicular 
to the opposite sides, they will all 
intersect at the point H. Join the 
points D, E, F. This will form a 
triangle of which H is the "in- 
centre," being the centre of the 
inscribed circle. The centres of 
the escribed circles will be the 
points A, B, and C. The radii of 
the circles are found by drawing 
lines from the centres perpendicular 
to the sides of the triangle produced (Prob. 7), 
line BK. 




Fig. 132. 



as the 



62 



GEOMETRICAL DRAWING AND DESIGN. 



95. 



To draw two circles tangential to three given straight lines, 
two of which are parallel. 



Let AB and CD be the two given 
parallel lines, and let the third line 
intersect them in E arid F. Bisect 
the four angles AEF, BEF, CFE, 
and DFE by lines meeting at H 
and L. From H draw the line HM 
perpendicular to CD (Prob. 7). 
With H and L as centres, and a 
radius equal to HM, draw the two 
required circles. 

Note. — A line joining H and L will be parallel to the two 
given lines AB and CD. 




Fig- 133- 



96. To draw two circles tangential to three given straight lines, 
none of which are parallel ; the third line to be drawn to cut 
the other two. 

Let AB, CD, and EF be the three given lines. Bisect the four 

j3 angles AFE, BFE, CEF, 
and DEF by lines meeting 
at H and L. From H and 
L draw the lines HM and 
LN perpendicular to CD 
(Prob. 7). With H as 
centre, and radius HM, 
draw a circle ; with L as 
centre, and LN as radius, 
draw the other circle re- 
quired. 

Note. — A line pro- 
duced through the points 
H and L would be the 

bisector of the angle formed by producing the Hues AB 

and CD. 




TANGENTS AND TANGENTIAL ARCS. 



63 



97. To draw DIRECT COMMON TANGENTS tO tWO given 

circles of unequal radii. 

Let AC be the radius of one circle, and BD of the other. 

Join the centres A and B. ,.--- --.^ 

From the centre A draw ~^!^^^^^=="r?==~— ____^ 
a circle with a radius /^ ^-j N^ '~~y>:^ — J- 

= AC-BD. Bisect the line / (^\~'\~Zr~~-r~Jr~'\\ 
AB at E (Prob. i). From I \^_'-J:,.y^'^^^ J 
E, with radius ExA., draw a \ ^, y _— — -7'^'^^* 

circle cutting the circle FKG ___--^^^^=-1p\ 

in the points F and G. Join '- -'' 

FB and GB. From F and B ^'-- ^35- 

draw the lines FO and BR perpendicular to FB (Prob. 7) ; 
and from the points G and B draw the Hnes GP and BS 
perpendicular to GB. Draw the line HL through the points 
O and R, and the line MN through the points P and S ; these 
will be the tangents required. 



98. To draw TRANSVERSE COMMON TANGENTS tO tWO given 
circles of unequal radii. 

Let A and B be the centres of the given circles, and AC and 
BD their radii. Join AB. 
With A as centre, and a 
radius = AC + BD, draw a 
circle. Bisect the line AB 
in E (Prob. i). With E as 
centre, and a radius equal 
to EA, draw a circle cutting 
the circle FKG in the points 
F and G. Join AF and AG, 
cutting ths given circle PCO 
in O and P. Join FB and GB. From B draw the line BS per- 
pendicular to FB (Prob. 7), also the line BR perpendicular to 
GB. Draw the Hne HL through the points O and S, and the 
hne MN through the points P and R ; these will be the 
tangents required. 




Fig. 1,6. 



64 



GEOMETRICAL DRAWING AND DESIGN. 



Tangential Circles and Arcs. 

99. Showing- the principle of tangential circles. 
One circle can touch another circle either internally or ex- 
ternally, and any number of circles can be drawn to touch a 

given line, as well as each 
other, in the same point. For 
instance take the point C on 
the given line AB. All circles 
that touch a given line in the 
same point touch each other at 
that point ; and all their centres 
will be on a line perpendicular 
to the given hne. 

If they are on the opposite 

sides of the given line, they will 

touch externally ; and if on the 

same side, will touch internally. 

If they are on the same side 

of the line, one circle will be 

entirely within the other. 

If their centres F and E are on the same side of the given 

line AB, the distance between them is equal to the difference 

of their radii ; but if their centres E and D are on the opposite 

sides of the given line, the distance between them is equal to the 

l' sum of their radii. 

__i.^_ The point of contact 

^ '' 'q " '^ >. ^ can always be found by 

joining their centres. 

100. To draw four equal 
circles with radius 
equal to given line D, 
with their centres on a 
given line AB ; two to 
touch externally and 
two internally a given 
_ circle, whose centre is 

I C and radius CG. 

13 ' With centre C, and 

Fig. 138. 





TANGENTS AND TANGENTIAL ARCS. 



65 



radius equal to the sum of the radii, i.e. CG + D, draw a circle 
cutting the given Hne AB in H and N. With centre C, and 
radius equal to the difference of the radii, i.e. CO - D, draw a 
circle cutting the given line AB in L and M. With H, L, M, 
and N as centres, and radius equal to D, draw the four circles 
required. 



101. To draw four equal 
circles, with radius 
equal to given line D, 
with their centres on a 
given arc AB ; two to 
touch externally and 
two internally a given 
circle, whose centre is 
C and radius CG. 

The construction of this 
problem is word for word 
the same as the last, the 
only difference being the 
words given arc instead of 
given line. 




Fig. 139. 



102 To describe a circle tan- 
gential to and including 
two given equal circles A 
and B, and touching one of 
them in a given point C. 

Find the centres of the two 
given circles D and E, and 
join them (Prob. 32). Join 
CD. From C draw the line 
CK parallel to DE, meeting 
the given circle B in K. Join 
KE, and produce it till it 
meets CD produced in F. 
With F as centre, and radius 
FC, draw the required circle. 




Fig. 40. 



66 



GEOMETRICAL DRAWING AND DESIGN. 



103. To describe a circle tangential to and including two unequal 

given circles A and B, 




Fig. 141. 



and touching one of them 

in a given point C. 

Find the centres D and 
E. Join CE. Cut off from 
CE, CH equal in length to 
the radius of the smaller 
circle. Join DH. Produce 
CE. At D construct the 
angle HDF equal to the 
angle DHF(Prob. 11). DF 
meets the line CE produced 
in F. With F as centre, 
and FC as radius, draw the 
circle required. 



104. To draw the arc of a circle having a radius of l:| inches, which 
. p shall be tangential to two 

\ given unequal circles A and 

B and include them. 
Note. — The diameters and 
distance between the circles 
must not be greater than 2^ 
inches. 

Find the centres D and E 
(Prob. 32), and produce a hne 
through them indefinitely in 
both directions, cutting the 
circles in K and L. From the 
points K and L on this line, set 
off KF and LH equal to the 
radius of the required arc, viz. 
I J inches. With D as centre, 
and a radius equal to DF, draw 
an arc at M ; and with E as 
centre, and EH as radius, draw 




Fig. 142. 



an arc intersecting the other arc at M. From M draw the 
line MD, and produce it till it meets the circumference of the 



TANGENTS AND TANGENTIAL ARCS. 



67 



larger circle in C. With M as centre, and MC as radius, draw 
the required arc. 

105. To inscribe in a segment of a circle, whose centre is E, two 

given equal circles with a radius equal to line D. 
From any radius EF cut off 
FL = D, and describe a circle with 
radius EL. Draw the line KL parallel 
to the base of the segment, and at a 
distance equal to given radius D from 
it (Prob. 4). Join EL and produce 
it till it meets the circumference in 
F. With L and K as centres, and 
LF as radius, draw the two required 
circles. 

106. In a given sector of a circle ABC, 
to inscribe a circle tangential to 
it. 

Bisect the angle ACB by the line CD 
(Prob. 12). At D draw a tangent HL 
(Prob. 84) to meet the sides of the 
sector produced. Bisect the angle 
CLH by a line cutting CD in E. 
With E as centre, and ED as radius, 
draw the circle required. ^ 

Fig. 144. 

107. Draw a circle having a radius of j of an inch tangential to 

two given unequal circles A and B externally. 

Note. — The circles must not be 
more than h inch apart. 

Find the centres D and E of the 
given circles (Prob. 32). From centre 
D, with the sum of the radii, z.e. 
D K + 1 of an inch, draw an arc at H ; 
and from centre E, with the sum of the 
other radii, z>. EL + j of an inch, draw 
another arc at H . With H as centre, and 
radius HK, draw the circle required. 

Fig. 145. 





68 



GEOMETRICAL DRAWING AND DESIGN. 




108. To draw the arc of a circle tangential to two given unequal 
circles A and B externally, and touch- 
ing one of them in a given point C. 

Find the centres D and E of the 
given circles (Prob. 32). Join CE, and 
produce it indefinitely. Set off from 

C, on CE produced, CH equal to the 
radius of the larger given circle. Join 
DH. At D construct an angle HDF, 
equal to the angle FHD, to meet EC 
produced in F. With F as centre, and 
radius FC, draw the arc required. 

109. To draw a circle, with a radius equal to given line C, tangential 
to two given unequal circles A and B, 
to touch A externally and B internally. 

Note. — The given radius must be 
greater than half the diameter of the 
enclosed circle and the distance 
between the circles. 

Find the centres D and E of the two 
given circles (Prob. 32). From centre 

D, with radius, the sum i.e. DH + C, 
describe an arc at F ; and with E as 
centre and radius the difference, i.e. 
C — EK, draw another arc cutting the 
FD and FE. With F as centre, and 




Fig. 147. 



Other at point F. Join 

FH as radius, draw the circle required. 



110. 



To draw a circle of f of an inch radius tangential to the 
given line AB and the given circle CDE. 

Note. — The circle must be less than 
I inch from the line. Draw the line 
KL parallel to AB, and f of an inch 
from it. Find the centre F of the 
given circle (Prob. 32), and with the 
sum-radius of the two circles draw the 
arc HM cutting KL in M. Join FM. 
With M as centre, and MN as radius, 
draw the required circle. 




TANGENTS AND TANGENTIAL ARCS. 69 



EXERCISES. 

1. Draw a circle 1.7" in diameter; at any point in its circumference, 
draw a tangent. 

2. Draw a circle 1.25" in radius ; from a point one inch outside the 
circle, draw a tangent to it. 

3. Draw a circle i|" in diameter ; from any point outside the circle, 
draw two tangents without using the centre. 

4. Draw two lines, enclosing an angle of 45° ; draw a circle i|" in 
diameter, tangential to these lines. 

5. At any point in the arc of a circle draw a tangent, without using 
the centre. 

6. Draw two circles of 2" and i" radii, with their centres 3" apart ; 
draw transverse common tangents to them. 

7. Draw two circles of 1.7" and i" radii, with their centres 2.75" 
apart ; draw direct common tangents to them. 

8. Draw two lines at an angle of 30° with each other, and a third 
line cutting them both at any convenient angle ; draw two circles 
tangential to all the three lines. 

9. Construct a triangle with sides 2.25", I.S", and 1.25"; draw an 
inscribed and three escribed circles tangential to the lines forming the 
sides. 

10. Draw two lines enclosing an angle of 45° ; fix a point in any 
convenient position between these two lines, and draw a circle that 
shall pass through this point and be tangential to the two lines. 

11. Draw a circle 1.25" in diameter, and half an inch from it draw 
a straight line ; draw a circle of §" radius, tangential to both the 
circle and the line. 

12. Draw two circles of i" and V' radius, with their centres 2h" 
apart ; draw another circle tangential to both externally. 

13. Draw two circles 1.50" and i" in diameter, their centres to be 
2.25" apart; draw another circle 3^" in diameter, touching the larger 
circle externally, and the smaller one internally. 

14. Draw two lines AB, AC, making an angle of 25° at A. Describe 
a circle off" radius touching AB and having its centre on AC. From 
A draw a second tangent to the circle, marking clearly the point of 
contact. (April, '96.) 

15. Draw a line AB, 2" long. Describe a circle of f " radius touching 
AB at A, and another of i" radius touching AB at B. Draw a second 
line which shall touch both circles, showing clearly the points of 
contact. (June, '97.) 

16. Describe two circles, each of h" radius, touching each other at a 
point A. Find a point B on one of the circles, t" from A. Describe a 
third circle, touching both the others and passing through B. (April, '98. ) 



7o 



GEOMETRICAL DRAWING AND DESIGN. 



17. Construct a square of 2" side. In the centre of the square place 
a circle of ^" radius. Then describe four other circles, each touching 
the first circle and two sides of the square. (June, '00. ) 

18. Draw the "cyma recta" moulding shown, (Fig. 149), adhering to 
the given dimensions. The curve is composed of two quarter-circles 
of equal radii, tangential to one another and to the lines AB and CD 
respectively. (April, '96.) 




Fig. i4q. 





Fig. 151. 



19. Draw the "Scotia" moulding shown (Fig. 150). The curve is 
made up of two quarter-circles of 1" and |" radius respectively. 

(May, '97.) 

20. Draw the "rosette" shown (Fig. 151), according to the figured 
dimensions. (June, '97). 

21. Draw the "ogee" arch shown (Fig. 152) to a scale of 2' to i". 
The arcs are all of 2' radius. The methods of finding the centres and 
points of contact must be clearly shown. (April, '99.) 






22. Draw the moulding shown (Fig. 153), adhering strictly to the 
figured dimensions. The arc of ^" radius is a quadrant. (April, '98.) 

23. Draw the "cyma recta" moulding shown in the diagram (Fig. 
154), using the figured dimensions. The curve is composed of two 

equal tangential arcs each of f" radius. (June, '98.) 

24. Describe a circle touching the given circle at 
T (Fig. 155) and passing through the point P (twice 
size of figure). (June, '99.) 

25. Describe the four given circles, using the 
figured radii (Fig. 156). The necessary construction 

lines and points of contact must be clearly shown. (June, '00.) 




^ig- 155- 



TANGENTS AND TANGENTIAL ARCS. 



71 



26. Draw the "three centred" arch shown (Fig. 157). The two 
lower arcs are of i" radius, and the upper arc is of 2^" radius. (April, '00. ) 




Fig. 156. 



Fig. 157- 



Fig. 158. 



27. Draw the "four centred" arch shown (Fig. 158), adhering to the 
figured dimensions. (June, '97.) 

28. Draw the figure shown, (Fig. 159), making tlie sides of the square 
21" long. (April, '96. ) 





*^ 




^ 


— - 3-^-"- - > 


iimi 


-i^^O^ 




Fig. 161. 




Fig. 160. 


Fig. 159- 



29. Draw the figure shown (Fig. 160), adhering strictly to the figured 
dimensions. (May, '97.) 

30. Draw the figure shown (Fig. 161), according to the figured 
dimensions. (June, '97.) 

31. Draw the given figure (Fig. 162) making the side of the square 
2^" and the radii of all the arcs f". (April, '99,) 



W- 



■£k 



2^' 

Fig. 162. 




Fig. 164. 



32, Draw the figure shown (Fig. 163), using the figured dimensions. 

(April, '98.) 

33. Copy the given figure (Fig. 164), using the figured dimensions. 

(June, '00.) 



CHAPTER VII. 



PROPORTIONAL LINES. 



Proportional lines may be illustrated by the example of simple 
proportion in arithmetic, in which we have four terms, e.g. 
2:4::5 : lo 

The relationship or ratio between the first two terms with 
regard to magnitude is the same as that between the second two, 
e.g. as 2 is to 4, so is 5 to ID ; therefore these four numbers 
are said to be in proportion. 

The first and fourth terms are called 
the extremes, and the second and third the 
means. 

The product of the extremes equals the 
product of the means, e.g. 2x10 = 4x5. 
So, the first three terms being given, we 
can find the fourth. If we divide the pro- 
duct of the means by the first extreme we 

get the fourth proportional, e.g. - — -= 10. 

Almost all geometrical questions on 
proportion are based on the following 
theorems : — 

Take any triangle ABC, and draw any 
line DE parallel to one side, then — 

CD :DA::CE:EB 
CD : CE : : CA : CB 
CE:ED::CB:BA 
CD : DE : : CA : AB. 
There are five varieties of proportional lines, viz. — 
Greater fourth proportional. 
Less fourth proportional. 
Greater third proportional. 
Less third proportionaL 
Mean proportionaL 




PROPORTIONAL LINES. 



73 



If the quantities be so arranged that the second term is 
greater than the first, — as 4 : 6 : : 8 :.v, — the last term is called 
the greater fourth proportional. 

If the terms are arranged so that the second term is less than 
the one preceding, — as 8 : 6 : : 4 : .r, — the last or unknown term 
is called the less fourth proportional. 

When the two means are represented by the same number, — 
thus 4 : 6 : : 6 : .r, — the answer or x is called the third propor- 
tional. 

The third proportional is found by dividing the square of the 

second by the first, e.£: — 

62 6x6 

— or = 9. 

4 4 

If the terms are placed so that the larger number is repeated, 
—thus 4:6::6:.i', — the last term is called the greater third 
proportional ; but if the terms are arranged so that the smaller 
number is repeated,— as 6 : 4 : : 4 : .r, — the result is called the 
less third proportional. 

The mean proportional between any two numbers is found by 
extracting the square root of their product, — e.g: 4x9 = 36 ; the 
square root of 36 = 6, which is the mean proportional. 



111. To find a fourth proportional to three given lines A, B, 

and C. THE GREATER FOURTH PROPORTIONAL. 

Draw EH equal to 
given line C, and EF, at 
any angle with it, equal to 
given line B. Join FH, 
and produce EH to D, 
making ED equal to given 
line A. Draw DK parallel 
to FH till it meets EF 
produced in K (Prob. 3). 
Then KE will be the 
greater fourth proportional 
to the lines A, B, and C, 
i.e. C : B : : A : KE, e.g. 
if C = 6feet, B = 8 feet, A = 




A. 

B: . 

C' ' 

Fig. 166. 

■ 12 feet, then KE=^ 16 feet 



74 



GEOMETRICAL DRAWING AND DESIGN. 



112. To find a fourth proportional to tliree given lines A, B, and C. 
THE LESS FOURTH PROPORTIONAL. 

Draw the line DE 
equal to given line A, 
and EF, at any angle 
with it, equal to given 
line B. Join FD. From 
E, along ED, set off 
EG equal to given line 
C. From G draw GH 
parallel to FD (Prob. 
3). Then HE is the 
less fourth proportional to 
the given lines A, B, and 
C, /.^. A : B : : C : HE. 




Fig. 167. 



113. To find a third proportional between two given lines A and B. 

THE GREATER THIRD PROPORTIONAL. 

Draw CD equal to given 
line A, and CE, at any 
angle with it, equal to 
given line B. Join DE. 
With C as centre, and CD 
as radius, draw the arc. DG 
to meet CE produced in 
G. From G draw the line 
GF parallel to DE till it 
meets CD produced in F 

^ (Prob. 3). Then CF is the 

greater third proportional 

' to the given lines A and B, 

Fig. 168. I.e. B : A : : A : CF. 




114, To find a third proportional between two given lines A and B. 

THE LESS THIRD PROPORTIONAL. 
Draw CD equal to the given line A, and CE, at any 



PROPORTION. 



75 



angle with it, equal to 
given line B. Join DE. 
From C as centre, and 
with radius CE, draw the 
arc EF cutting CD in F. 
Draw FG parallel tc DE 
(Prob. 3). Then CG is 
the less third proportional 
to the given Hnes A and B, 
z\e.A:B::B: CG. 



\ 1 


\ 


^-v^ 




G 


C 





Fig. 169. 

115. To find the MEAN PROPORTIONAL between two given lines 
AB and CD. 

Produce the given line AB 
to E, making AE equal to 
the given line CD. Bisect /^ 

the line EB in H (Prob. i). 
From H as centre, and with / 
radius HB, draw the semi- / 
circle EKB. At A draw the / 
line AK perpendicular to EB, |~ 
cutting the semicircle in K 

(Prob. 5). Then AK is the 1 

mean proportional to the given ^ 

lines AB and CD, e.o-. if 

AB = 9 feet and CD -4 feet, then AK: 



H 



A 



Fie. 



6 feet. 



116. 



To divide a line in medial section, i.e. into EXTREME and 
MEAN proportion. 



Let AB be the line. At A 
draw AC perpendicular to 
AB, and equal to it. Bisect 
AC in D. With D as centre, 
and DB as radius, draw the 
arc cutting CA produced in 
E. With A as centre, and 
AE as radius, draw the arc 
cutting AB in F. Then 
AB : AF : : AF : BF. 



B 



76 



GEOMETRICAL DRAWING AND DESIGN. 



117. To divide any straight line AB in the point C, so that 
AC : CB : : 3 : 4. 




At A draw the line AD of 
indefinite length, and at any 
angle to AB. From A, along 
AD, mark off seven equal 
distances of any convenient 
length. Join 7B. At 3 draw 
the line 3C parallel to 7B 
(Prob. 3). Then 

AC : CB : : 3 : 4, 
or, AB : AC : : 7 : 3. 



118. To divide a line proportionally to a given divided line. 



Let AB be a given divided 
line, and GH a line which is 
to be divided proportionally 
to AB. Join GA and HB, 
and produce them to meet in 
C ; join C to the divisions 
in AB and produce them. 
These lines will divide GH 
as required. Divide BC into 
four equal parts, and draw 
the lines D, E, and F parallel 
to AB ; join the divisions on 
AB with C, then the divisions 
on the lines D, E, and F will 
represent respectively the 
proportions of |, ^, and j 
of the divisions on the given 
line AB. 



/ / ' , 

f • > 

^ / / i — i 
\^ — / — f '. 

_/ / hJ 



PROPORTIONAL LINES. 



77 



119. To construct a triangle on a given line AB, so that the 
three angles may he in the proportion of 2 : 3 : 4. 

From B, with any radius, describe a semicircle and divide it 
into nine equal parts (Prob. 48). Draw the lines B4 and B7. 
Then the three angles 7BC, 
4B7, and 4BA are in the 
proportion of 2 : 3 : 4. The 
sum of the three angles are 
equal to two right angles, 
because a semicircle con- 
tains 180° ; so they must 
be the three angles of a 
triangle, because the three 
angles of any triangle are 

together equal to two right angles. From A draw the line 
AD parallel to B7 till it meets B4 produced in D (Prob. 3). 
Then ABD is the triangle required. 




120. This problem illustrates an important principle in proportion. 

Take a triangle ABC, the sides of which shall bear a 
certain ratio. For example, 

let AB : BC as 2 : i. Pro- D 

duce AB to D, and bisect the / ^, 

angles ABC and DBC by 
lines meeting AQ and AC 
produced in H and E. 
Bisect the line HE in K 
(Prob. i). With K as centre, 
and KE as radius, draw the 
circle EBH. Now, if we 
take any point M in this 
circle, and join MA and MC, 
we shall find that they bear 
the same ratio as the lines 
AB and BC. In the example 
given MA : MC as 2:1. The same result would be obtained 
from any point in the circle. 




78 



GEOMETRICAL DRAWING AND DESIGN. 



121. To divide a right angle into five equal parts. 




Let ABC be the right 
angle. Divide BC in D, so 
that BC : BD : : BD : DC 
(Prob. 1 1 6). With C as 
centre, and CB as radius, 
describe the arc BE ; and 
with B as centre, and BD as 
radius, describe the quadrant 
DF, cutting BE in E FE is 
one-fifth of the quadrant FD. 
Arcs equal to it set off on 
FD will divide it into five 
equal parts. 



122 



--— ^.E 



To find the Arithmetic, the Geometric, and the Harmonic means 
between two given lines AB and BC. 

Bisect AC in D (Prob. i). 

With D as centre, and radius 

DA, draw the semicircle 

^ AEC. At B draw BE per- 

'^ pendicular to AC (Prob. 5). 

\ Join DE. From B draw BF 

I perpendicular to DE. AD 

D B C. is the Arithmetic, BE the 

Geometric, and EF the 




177. 



Harmonic mean between the two lines as required. 




Fig. 178. 



123. Taking the given line AB as the unit ; find 
lines representing \^ and sfs. 

Draw AC perpendicular to AB, and of the 
same length (Prob. 5). Join i:B. CB = n/2. 

Draw CD perpendicular to CB, and equal 
in length to AB and AC. Join DB. DB = v'3. 

If AB is the edge of a cube, CB is the 
diagonal of its face, and DB the diagonal of 
the cube, which are therefore to one another 
as I : \/2 : \/3. 



PROPORTIONAL LINES. 



79 



Table of Foreign Road Measures and their 
Equivalents in English Yards. 







English Yards. 


Austria, - - • 


mile 


8297 


Bavaria, - 


)) 


8059 


Belgium, - - - - 


kilometre 


1094 


Berne, - - - - 


league 


5770 


China, . = - - 


li 


609 


Denmark, 


mile 


8238 


England, - - = - 


55 


1760 


France, - - - - 


kilometre 


1094 


Germany, 


mile 


8101 


Greece, - - - - 


)) 


1640 


Holland, - - - - 


5» 


1094 


India (Bengal), 


coss 


2000 


Italy, - - - - 


mile 


2025 


Netherlands, - 


kilometre 


1094 


Norway, - - - - 


mile 


12,182 


Persia, . - . - 


parasang 


6076 


Portugal, - - - - 


mile 


2250 


Prussia, - - - - 


5> 


8238 


Russia, - - - - 


verst 


I167 


Siam, ... - 


roeneng 


4204 


Spain, ... - 


mile 


1522 


Sweden, . - - - 


55 


11,690 


Turkey, - - - - 


berri 


1827 



EXERCISES. 

1. Draw three lines 1.25", 2.3", and 2.75" respectively, and find 
their greater fourth proportional. 

2. Draw two lines 2.7" and 1.5" in length, and find their less third 
proportional. 

3. Draw a line 2.5" in length, and produce it so that its extra length 
shall be in proportion to its original length as 3 : 5. 

4. Draw two line's 25" and |" in length, and find their mean 
proportional. 

5. Construct a triangle on a base 3.25" in length, so that its three 
angles are in the proportion of 3, 4, and 5. 



8o GEOMETRICAL DRAWING AND DESIGN. 

6. Divide a line 3.4" in Kngth in extreme and mean proportion. 

7. Divide a line 2.75" in length so that one part is in proportion to 
the other as 2 : 4. 

8. Draw two lines 1.25" and 2.3" respectively, and find their greater 
third proportional. 

9. Draw three lines 3", 2^", and i|" in length, and find their less 
fourth proportional. 

10. The base of a triangle is l '6^" long, and the angles at the base 
are 88° and 53°. Construct the triangle, and find a fourth proportional 
less to the three sides. Measure and write down the length of this line. 
(The angles should be found from the protractor or scale of chords.) 

(April, '01.) 



CHAPTER VIII. 

PLAIN SCALES, COMPARATIVE SCALES, AND 
DIAGONAL SCALES. 

On a drawing representing a piece of machinery the scale is 
written thus : Scale j full size. From this we know that every 
inch on the drawing represents 4 inches on the actual machine, 
so the relation between any part represented on the drawing and 
a corresponding part in the real object is as i : 4 or j. This is 
called the representative fraction. 

A drawing representing a building has drawn upon it a scale; 
e.g. — Scale \ of an inch to a foot. One-quarter of an inch is 
contained forty-eight times in i foot, so the R.F. is -^^. 

On a large drawing showing a district the scale is written 
thus : R.F. yyVo- As there are 1760 yards to a mile, it is 
evident that every 3 feet on the drawing is equal to i mile 
on the land represented. This, of course, is a very large 
scale. 

Our Ordnance Survey Office publishes a map of 25 inches to a mile, 
which is useful for small districts or estates ; one of 6 inches to a mile, 
useful for maps of parishes ; and one of I inch to a mile, useful for 
general purposes. 

The R.F. for the last would be g^eo- 

mile, yards. feet. inches. 
I = 1760=5280 = 63,360. 
F 



82 



GEOMETRICAL DRAWING AND DESIGN. 



124. 



To construct a scale 4 inches long, showing inches and 
tenths of an inch. Fig. 179. 

Draw a line 4 inches long, and 
divide it into four equal parts, each 
of which will be i inch. At the end 
of the first inch mark the zero point, 
and from this point mark the inches 
to the right i, 2, 3. These are called 
primary divisions, and the amount 
by which they increase is called the 
value of the scale length. 

The division left of the zero 
point has to be divided into ten 
equal parts. The best way to do 
this is to take a piece of paper and 
set off along its edge ten equal 
divisions of any convenient size 
(Fig. 179). Produce the perpendi- 
cular marking the division at the 
zero point, and arrange this piece 
of paper so as to fit in exactly 
between the end of the division and 
this perpendicular line. If we now 
draw lines parallel to the perpendi- 
cular at the zero point, they will 
divide the inch into ten equal parts. 
These are called the secondary 
divisions ; they have the same zero 
point as the primary, and their 
numbers increase from this poini 
by the value of their scale length. 

This scale will measure inches 
to one decimal place. Supposing 
we wish to measure 3.6 inches, 
that is 3 primary and 6 secondary 
divisions. Place one point of the 



\ 


.\X V 

- ro -^ 






1- 

Co 





Fig. 179. 



dividers on division 3 of the primary parts, and open them 
till the other point reaches the division (^ of the secondary 
divisions. 



PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 8- 



125. 



<T>- 



00- 



cs- 



lO' 



HI? 



^-- 



to- 



CVJ 



To construct a scale of g^g, or 1 incli to equal 3 feet 

Fig. i8o. 
Number of feet to be repre- 
sented may be assumed at 
pleasure, say I2 feet. 

36 : 12 : : 12 :;r, 
I2X 12 



whence ,r: 



4 inches. 



Fig. 180. 



36 

Draw a Hne 4 inches long, 
and mark off each inch. Trisect 
each of these divisions by a piece 
of paper, as shown in Fig. 179. 
We now have the total length 
divided into twelve ecjual parts. 

At the end of the first division 

*§ mark the zero point, and from 

^•^ this point towards the right, 

^ figure the primary divisions i, 

2, 3, etc. : these will represent 

feet. To the left of the zero 

point mark oflf twelve divisions : 

these will represent inches. 

In this scale the numbers 
increase on each side of the 
zero point by unity. 

This scale will measure feet 
and inches. 

126. To construct a scale with the 
R.F. o|s, or 1 inch to equal 
8 yards ; to measure 40 yards. 

Fig. 18 T. 

288 : 40 : : 36 :;r, 
36x40 



whence x 



288 



= 5 mches. 



O 



Draw a line 5 inches in length, and divide it 
into four equal parts, i.e. 1.25 for each division. 
At the end of the first division mark the zero 
point. As each of the primary divisions is 



O- 

00 

O. 
Fig. 181 



I 



84 



GEOMETRICAL DRAWING AND DESIGN. 



ro-- 



C\] 



O 

c\J 

CD 

Ik 

Fig. 182. 



^ 
^ 



equal to 10 yards, we must 
figure them from the zero 
point to the right 10, 20, 
and 30 yards. 

The division to the left 
of the zero point we divide 
into ten equal divisions. 
Each of these secondary 
divisions represents i 
yard. 

127. To construct a scale, 
RF. 6 356 o> or 1 iiich 
to a mile ; to measure 
5 miles. Fig. 182. 

If I mile=i inch, 5 
miles = 5 inches. 

Draw a line 5 inches 
long. Mark off each inch. 
At the end of the first 
division mark the zero 
point, and number the 
primary divisions to the 
right I, 2, 3, 4 miles. 
Divide the division to the 
left of the zero point into 
eight equal parts : these 
secondary divisions repre- 
sent furlongs. 

Comparative Scales. 



o 



0-- 



0J-- 



On an old French map 
a scale of leagues is -^ 
shown, as Fig. 183. Upon 
measuring this scale with qo 
an English scale, 30 
leagues are found to coin- oo 
cide with 4 inches. 



o 

00 



I 






o __ 

CM 



o ■ 

CO 

00 
O - 



Fig. i»3. 



PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 85 



128. To construct 
Scale to 
A French league = 
French 

leagues, miles, 

4262.84 



a comparative scale of English miles 
measure 80 miles. Fig. 183. 
4262.84 English yards. 
English O 



30 = 



1760 



X 30 



4262.84 X 30 



1 760 
whence 

4 X 80 X 1 760 



: 80 : : 4 : x 



4.4 inches nearly. 



4262.84x30 

Draw a line of this length, and place 
the zero point at the left-hand end and 
80 at the other extremity. Divide this 
line into eight equal divisions : each 
of these primary divisions will repre- 
sent 10 miles. 

For the secondary divisions, set off 
one of the primary divisions to the left 
of the zero point, and divide it into ten 
equal divisions : each of these will 
represent i mile. The representative 
fraction of both the French and English 
scales will of course be the same. 

On a Russian map a scale of versts 
is shown, as Fig. 184, by measuring 
which by an English scale 
120 versts = 4 inches. 

129. To construct a comparative scale of 
English miles Scale to measure 80 
miles. Fig. 184. 

A Russian verst=ii67 English yards. 
1 167 X 120 



120 versts = 



1 167 X 120 



1760 



miles. 



1760 



80 : : 4 : X, 



o— - 



I 



o 

00 



whence 



4 X 80 X 1 760 
1167x120 



= 4 inches nearly. 



R 



o 



o 






o 

CO 



o 

CM 



o- 

CVJ 

to 

00 

o_ 



-^ 



Fig. 184. 



86 GEOMETRICAL DRAWING AND DESIGN. 

Draw a line of this length, and divide it into eight equal 
divisions : each of these primary divisions will represent lo miles. 
Place the zero point at the left-hand end of the line, and figure 
the divisions towards the right lo, 20, 30, etc. Set off one of 
the primary divisions to the left of the zero point, and divide it 
into ten equal divisions : each of these will represent i mile. 

Diagonal Scales. 

In the preceding scales we have only primary and secondary 
divisions, and if we wish to measure a fractional proportion of a 
secondary division, we cannot do it with any accuracy ; but by 
means of a diagonal scale we are enabled to measure hundredths 
of primary divisions, as will be seen from the following scale. 

130. To construct a diagonal scale 3 inches long, to measure inches, 
tenths of inches, and hundredths of inches. Fig. 185a. 
Draw a rectangle ABDC 6 inches long and about ij inches 
wide, and divide it into six equal parts. At the end of the first 



I" I I I I I 



Fig. i«5a. 



Fig. 185b. 



N.B. — These two figures are half the size described in the text, and should be 
drawn full size by the student. 

division from A fix the zero point, and to the right of this 
figure each division i, 2, 3, 4, and 5. Divide AC into ten 
equal parts, and figure them from A towards C, then draw lines 
parallel to AB from one end of the scale to the other. Divide 



PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 87 

Ao into ten equal divisions, and figure them from o towards A. 
Join 9 to C, and from each of the other divisions between A 
and o draw hnes parallel to 9C. Note. — The divisions between 
o and B are primary, between o and A secondary and between 
A and C tertiary. 

To take off from this scale a measurement equal to 2.73 
inches, we place one point of the dividers on the primary 
division figured 2, and the other on the secondary division 
figured 7, but both points must be on the line that is figured 3 
on AC. The points are marked by a small circle on the scale. 

To take off 3.17 inches, place one point of the dividers on 
the primary division 3, and the other on the intersection between 
the secondary division i and the line 7 on AC. These points 
are shown by two crosses on the scale. 

131. To construct a diagonal scale showing miles, furlongs, and 
chains, to show 6 miles, R.F, = g3.^gQ. P^ig. j85b. 

I mile = 8 furlongs. 
T furlong=io chains. 
The length of scale = g3^go of 6 miles = 6 inches. 

In this scale there will be six primary, eight secondary, and 
ten tertiary divisions. 

Construct a rectangle ABDC 6 inches long and about ij 
inches wide, and divide it into six primary divisions. Place the 
zero point o at the end of the first division from A, and divide 
Ao into eight secondary divisions, figured from o towards A. 
Divide AC into ten equal divisions, and figure them from A to 
C. Join the secondary division figured 7 to the point C, and 
from each of the other secondary divisions draw lines parallel 
to 7C, thus completing the scale. 

To take off from this scale 2 miles, 5 furlongs, and 7 chains, 
take one point on the primary division 2, and the other where 
the line from the secondary division 5 intersects the division 
7 on AC. These two points are marked by two small circles on 
the scale. 

To take off 3 miles, 2 furlongs, and 3 chains, one point will be 
on the primary division 3, and the other where the secondary 
division 2 intersects the tertiary division 3. These points are 
marked by two small crosses on the scale. 



88 



GEOMETRICAL DRAWING AND DESIGN. 



132. To take off any number to three places of figures from 
a diagonal scale. 

On the parallel indicated by the third figure, measure from the 
diagonal indicated by the second figure 
to the vertical line indicated by the first. 



DISTANCE 

61 




lOLOMtrJlES 

Fig. 186. 



Comparative Diagonal Scales 

useful for transferring the 
value of quantities in one 
measure to another. To 
make the scales less cum- 
bersome they are so 
arranged that in some 
cases the number to be 
taken off must be halved. 
With this proviso, any 
quantity may be converted 
from one scale to another. 
The number expressing 
the quantity in one unit is 
taken off on the scale for 
that unit, and the number 
expressing it in the other 
unit is at once read off on 
the parallel scale. 

For example, a length 
of 638 miles. Its half, 
319 miles, corresponds to 
5 1 3 kilometres, so that 638 
miles corresponds to 1026 
kilometres. 



The diagonal scale 
generally found in instru- 
ment-boxes is shown in 
Fig. 187. 
diagonal scales. In one, 



are very 



3) y> 


«-N 






7> 






>o 


f 




<t> 






^-■■U 


■■■«» 















- 


■ f 






4- . 


■ rt 


: , II 




1 ty 










_ 




~ 


- 


CD 




- T 












- i - ■ 


•r" 


:i:,: 




:::|- 


!:::" 


eoo >■ 


el 



Fig. 187. 

It consists of two diagonal scales. In one, the distance 
between the primary divisions is half an inch, and in the other 
a quarter of an inch. 



PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 89 

There is a small margin on each side of the scale for figures : 
on one side the half inches are figured, and the quarter inches 
on the other. 

One primary division at each end is divided into ten secondary 
divisions, and there are ten tertiary divisions drawn from one 
end of the scale to the other. 

The primary divisions bemg taken for units, to set off the 
numbers 5.36 by the diagonal scale. This measurement is 
shown by two crosses on the scale. 

If we reckon the primary divisions to stand for tens, the 
dimension would have one place of decimals, e.g. to take off 36.4 
from the diagonal scale. These points are shown on the scale 
by two small circles. 

The primary divisions being hundreds, to take off 227. This 
dimension is shown on the scale by two small squares. 



Proportional Scales. 

These are used for enlarging or reducing a drawing in a given 
proportion : three varieties are here illustrated. 

The simplest form is that shown in Fig. 188. Suppose we wish 
to enlarge a drawing in the proportion of 3 : i. 

Draw the line AB of convenient 
size, to suit the measurements on 
the drawing, and produce it to C ; 
make BC one-third of AB. On AB 
erect the perpendicular BD any 
length, and join AD and DC. 
Divide BD into any number of 
equal parts, and draw lines parallel 
to AB. These lines are simply a 
guide to enable the measurements 
to be made parallel to the base — 
e.g. on placing a measurement from Fig. 188. 

the original drawing on the scale 

we find it occupies the position of ef: the distance between e 
and g will then give the length of the measurement to the 
enlarged scale, i.e. in the proportion of 3 : i. We proceed in the 
same manner with every measurement we wish to enlarge. 





D 




\ 


/- 


~~\ 


./ 


e V 


/ : 




/ 


\ 


A 


i c 



@ 


■■•••. 




^--. / 




F ty ^ 


/ 


-- — ;>%. 


/rt 




/9 


I 


V' 



90 GEOMETRICAL DRAWING AND DESIGN. 

Should we wish to reduce a drawing in the same proportion, 
viz. I : 3, the original measurements would be placed on the 
left-hand side of the scale, and the required proportion taken 
from the right-hand side. 

In Fig. 189 we have a series of measurements — Az', Kh^ Ag^ 

etc. — which we wish to enlarge, say in the proportion of 3 : 2. 

^ Draw the line AB any convenient 

length to suit size of drawing. 

From B draw BD perpendicular 

to AB. Produce AB to C, and 

make BC equal to half of AB. 

With centre A, and radius AC, 

draw an arc till it meets BD in D ; 

and join DA. From each of the 

points, /, //, g, /, etc., draw Hnes 

parallel to BD. The distances A/', 

Ak\ A^, etc., will then give the 

original measurements to the en- 

^ Fig. 189. larged scale of 3 : 2. 

To reduce the original drawing 
in the same proportion, i.e. 2 : 3. With A as centre, and 
radius AB, draw the arc BE. From E draw the line EF 
parallel to BD. AF will then represent AB reduced in the 
proportion of 2 : 3, and so on with any other measurement that 
we may require. 

EXERCISES. 

1. Construct a scale to measure feet and inches; the R.F. to be tjVj 
and its scale length value 15 feet. 

2. Construct a scale to measure yards and feet, the R.F. to be ^V, 
to measure 18 yards. 

3. Construct a diagonal scale to measure feet and inches, R.F. yV, 
to measure 36 feet. Take off a length of 17' 9". 

4. On a map, a distance known to be 20 miles measures 10" ; 
construct a diagonal scale to measure miles and furlongs, long enough 
to measure 12 miles. 

5. Construct a diagonal scale to measure yards and feet, R. F. ■^\-^, 
to measure 30 yards. 

6. On a map showing a scale of kilometres, 60 are found to equal 
3". What is the R.F.? Construct a comparative scale of English 
miles, to measure 100 miles. 



PLAIN, COMPARATIVE, AND DIAGONAL SCALES. 91 

7. A line 4f " long represents a distance of 4'. Construct a scale by 
which feet and inches may be measured up to 4 feet. The scale must 
be neatly finished and correctly figured. (April, '96.) 

8. Construct a scale to show yards and feet, on which 3^" represent 
8 yards. Make the scale long enough to measure 10 yards, and finish 
and figure it properly. (May, '97.) 

9. Construct a scale of 7^' to i", by which single feet may be 
measured up to 30'. The scale must be neatly finished and correctly 
figured. (June, '97.) 

10. The diagram represents an incomplete scale of feet (Fig. 190.) 
Complete the scale so that distances of 2' may be 

measured by it up to 50'. The scale must be ' ^ J^ 

properly finished and figured. (April, '98. ) Fig. 190. 

11. Construct a scale one-tenth of full size, to measure feet and 
inches up to 5 feet. The scale must be properly finished and figured. 

(June, '98.) 

12. Construct a scale of feet and inches one-ninth ( ^) of full size long 
enough to measure 4 ft. The scale must be properly finished and 
figured, and should not be "fully divided" throughout i.e. only one 
distance representing i ft. should be divided to show inches. (April, '99. ) 

13. T he line AB (2.75 inches long) represents a distance of 2^ ft. 
Make a scale by which feet and tenths of a foot may be measured up 
to 4 ft. The scale must be properly finished and figured, and should 
not be "fully divided " throughout, ?'.<?. only one distance representing 
I foot should be divided to show tenths. (June, '99.) 

14. The given line AB (3.2 inches long) represents a distance of 35 ft. 
Construct a scale by which single feet may be measured up to 40 ft. 
The scale is not to be "fully divided" i.e. single feet are not to be 
shown throughout the whole length, and it must be properly finished 
and figured, (April, '00. ) 

15. A drawing is made to a scale of i^" to i' , and another drawing 
is required on which the dimensions shall be three-quarters of those on 
the fir.-t drawing. Make a scale for the second drawing to show feet 
and inches up to 5', The scale is not to be "fully divided" [i.e. only 
one length of i' is to be divided to show inches) and it must be properly 
finished and figured. (June, '00.) 

16. Construct a scale \ of full size, by which feet and inches may be 
measured up to 2 feet. Show also distances of ^" by the diagonal 
method. Finish and figure the scale properly, and show by two small 
marks on it how you would take off a distance of i' 7V'. (April, '01.) 

17. Draw a diagonal scale, one-tenth of full size, by which centi- 
metres may be measured up to one metre. Mark on the scale a length 
of 47 centimetres. 

One metre (100 centimetres) may be taken as 39". (^lay, '03,) 



CHAPTER IX. 

INSTRUMENTS FOR MEASURING ANGLES, ETC. 

A protractor is an instrument used for measuring or setting off 
angles ; it may be either semicircular or rectangular in shape, as 




shown in Fig. 191. The point C marks the centre from which 
the radiating lines are drawn, and corresponds with the centre 
of the circle. 

The degrees are numbered in primary divisions, equal to 
ten degrees each, on the outside line from A; and on the inside 
line from B. In the actual instrument each of these primary 
divisions is subdivided into ten secondary divisions, each of 
which represents one degree. Only one of these is divided in 
the figure. 



133. To construct a scale of chords. 

A scale of chords is constructed in the following manner 
(Fig. 192). Draw the Hues AC and CD perpendicular to each 



INSTRUMENTS FOR MEASURING ANGLES, ETC. 93 

other. With C as centre, draw any quadrant AED, and 
divide the arc into degrees (only the primary divisions are 
shown in the figure). Join AD. With A as centre, and each of 
the primary divisions as radii, draw arcs cutting the chord AD, 
which will form the scale 
of chords. . ??- 

To use this scale in 
setting off an angle — for 
example, to draw a line 
that will make an angle 
of 40° with line CB (Fig. 
192). 

With C as centre, and 
radius equal to A60 on 
the scale of chords, draw an arc BFD. With a pair of dividers, 
take the distance A40 from the scale, and set it off on the arc 
BF from B. Join FC. Then FCB will be the angle of 40° 
required. 

Note. — A60 is always equal to the radius of the quadrant 
from which the scale of chords is constructed. 




134. To construct any angle without a protractor. 

Draw CD perpendicular to AB. With C as centre, and 
CA as radius, draw the semicircle ADB. Trisect the angle 
DCB in E and F (Prob. 13). Trisect the angle ECB in H and 
K (Prob. 14). Bisect FK 
in L (Prob. 12). Then 
DE = 3o°,EH = 2o°,HF=io°, 
and FL = 5°. Therefore 
between D and B we can 
construct any angle that is a 
multiple of 5°. 

Divide the angle ACD into 
five equal parts by the radii 
from M, N, O, and P (Prob. 121). From A set off AR equal to 
DE. As AN = 36° and AR = 3o°, AN-AR = 6°, MR = 12°, 
AM = 18°, and RO = 24°. Therefore between A and D we can 
construct any angle that is a multiple of 6°, 




tig. 193. 



94 GEOMETRICAL DRAWING AND DESIGN. 

If we subtract the multiples of 5° from those of 6° we can 
obtain any desired angle, e.g. — 

6- 5 = 1° 

12-10 = 2° 

18-15 = 3° 
24-20 = 4° 

30-25 = 5° 

etc. etc. etc. 

All the regular polygons, with the exception of two — the 
heptagon and undecagon — can be constructed with angles that 
are multiples of 5° or 6°. 

If the polygon is to be inscribed in a circle, the angle would 
be set off at the centre of the circle ; but if one side of the 
polygon is given, the angle would be set off externally, as 
shown in Fig. 74, page 31. 

The exterior angle of a Pentagon is 72° a multiple of 6°. 
„ „ Hexagon „ 60° „ 6°. 

„ „ an Octagon „ 45° „ 5°. 

„ „ a Nonagon „ 40° „ 5°. 

M i' Decagon „ 36° „ 6°. 

„ „ Duodecagon „ 30° „ 5°and6°. 



The Sector. 

The Sector is an instrument of great utility for facihtating the 
work of Practical Geometry. It consists of graduations on the 
two radii of a foot-rule, and it is used by measuring the arc 
between the graduations. Hence its name. The legs can be 
opened to contain any angle up to a straight line. 

In the illustration (Fig. 194) only the lines most used in 
Practical Geometry are shown : viz. line of lines, marked L on 
each leg ; a pair of lines of chords, marked C ; and a line of 
polygons, m.arked POL, on the inner side of each leg. 

The sectoral lines proceed in pairs from the centre of the 
hinge along each leg, and although the scales consist of two 
or three lines, parallel to the sectoral lines, all measurements 
must be made on the inner lines of each scale, i.e. the lines 
that radiate from the centre. 



INSTRUMENTS FOR MEASURING ANGLES, ETC. 95 



When the measurement is confined to a Hne on one leg of the 
sector, it is called a lateral distance ; but when it is taken from a 




Fig. 194. 

point on a line on one leg to a similar point on a corresponding- 
line on the opposite leg, it is called a transverse distance. 

Simple proportion. — Let AB and AC (Fig. 195) represent 
a pair of sectoral lines, and BC and DE b, 
two transverse measurements taken between 
this pair of Hnes ; then AB is equal to AC, 

and AD to AE, so that AB : AC : : AD : AE, d\ - - -7£ 

and the lines AB : BC : : AD : DE. 



The Line of Lines. 

The primary divisions only are shown in 
the illustration ; in the real instrument, each 
of these is subdivided into ten secondarv divisions 




135. To find the fourth proportional to three given lines. 

From the centre, measure along one leg a lateral distance equal 
to the first term ; then open the sector till the transverse distance 



96 GEOMETRICAL DRAWING AND DESIGN. 

between this point and a corresponding point on the other leg 
is equal to the second term ; then measure from the centre 
along one leg a lateral distance equal to the third term ; the 
transverse distance from this point to a corresponding point on 
the opposite leg will then give the fourth term. 

Example. — To find the fourth proportional to the numbers 3, 
4 and 9. From the division marked 3, which is the first term, 
open the sector till the distance between this point and the 
corresponding division on the other leg is equal to 4 divisions : 
this will be the second term. Then 9 being the third term, the 
transverse distance between the corresponding divisions at this 
point will give the fourth term, viz. 12. 



136. To find the third proportional to two ^ven lines or 
numbers. 

Make a third term equal to the second, then the fourth term 
will give the required result 



137. To bisect a given line. 

Open the sector till the transverse distance between the end 
divisions, 10 and 10, is equal to the given line ; then the transverse 
distance between 5 and 5 will bisect the given Hne. 



138. To divide a given line AB into any number of equal parts. 

For example, eight (Fig. 196). When the number of parts is a 

power of 2, the division 

A ' D — -^ Q ' ' ' ^ is best performed by suc- 

Fig. 196. cessive bisections. Thus, 

make AB a transverse 
distance between 10 and 10, then the distance between 5 and 
5 will give AC = half AB. Then make the transverse distance 
between 10 and io = AC, the distance between 5 and 5 will 
then give AD = one quarter of AB. By repeating the operation 
each quarter will be bisected, and the given line divided into 
eight equal parts as required. 



INSTRUMENTS FOR MEASURING ANGLES, ETC. 97 

When the number of divisions are unequal,— for example, 
seven (Fig. 197), — make the transverse distance between 7 and 

I 1 I I 1 I ' I 

A c B 

Fig. 197. 

7 equal to the given line AB ; then take the distance between 
6 and 6, which will give AC. The distance CB will then divide 
the line into seven equal parts. 

139. How to use the sector as a scale. 

Example. — A scale of 1 inch equals 5 chains. Take one inch 
on the dividers, and open the sector till this forms a transverse 
distance between 5 and 5 on each line of lines ; then the corre- 
sponding distances between the other divisions and subdivisions 
will represent the number of chains and links indicated by 
these divisions : for instance, the distance between 4 and 4 
represents 4 chains, 6.5 = 6 chains 50 links, Z-7="3 chains 70 
links, etc. 

Note. — i chain is equal to 100 links. 

140. To construct a scale of feet and inches, in which 2^ inches shall 
represent 20 inches. 

Make the transverse distance between 10 and 10 equal to 2\ 
inches ; then the distance between 6 and 6 will represent 12 inches. 
Make AB (Fig. 198) equal to this length. Bisect this distance 

' P . . ^ . t . , ^ . t ^ / ^.^ 

A D C E F B " 

Fig. 198. 

in C, as described for Fig. 196 ; then bisect AC and CB in D 
and E in the same manner. Take the transverse distance 
between 5 and 5, which will give AF 10 inches ; EF will then 
trisect each of the four divisions already obtained. AB will then 
be divided into twelve divisions, which will represent inches. 
Produce the line AB to H, and make BH equal to AB. BH 
will then represent one foot. 

G 



98 GEOMETRICAL DRAWING AND DESIGN. 

141. How the sector may be used for enlarging or reducing a 
drawing. 

Let ABC (Fig. 199) represent three points in a drawing, let 
it be required to reduce this in the proportion of 4 to 7. 
Make the transverse distance between 7 and 7 equal to AB ; 
then take the distance between 4 and 4, and make DE equal to 
this length. Also make the distance between 7 and 7 equal to 

A 




Fig. 199. 

AC ; then take the distance between 4 and 4, and from D as 
centre, with this distance as radius, describe an arc. In the 
same manner make the distance between 7 and 7 equal to BC ; 
then with a radius equal to 4, 4, describe another arc from E, 
cutting the other arc in F. Join EF and DF. Then DEF will 
be a reduced copy of ABC, in the proportion of 4 : 7 as required. 

142. To enlarge a drawing in the proportion of 7 to 4. 

In this instance the sector would be opened so that the 
transverse distance between 4 and 4 should represent the 
original measurements, while those required for the copy would 
be taken between 7 and 7. 

The Line of Chords. 

In the scale of chords already described (Prob. 133) we are 
limited to one radius in setting off angles — viz. a radius equal to 
the 60 marked on the scale ; in the double line of chords on the 



INSTRUMENTS FOR MEASURING ANGLES, ETC. 99 



sector there is no such hmitation — we can set off any radius 
equal to the transverse distance between the two points 60 and 
60, from their nearest approach to each other up to the fullest 
extent the opening of the sector will admit of. 



143. To construct an angle of 50^. 
Open the sector at any convenient distance. 



Take the 




transverse distance between the 
points 60 and 60, and construct an 
arc with this radius. Let AB 
(Fig. 200) represent this radius. 
Now take the transverse distance 
between 50 and 50, and set it off 
from B on the arc, which will give 
the point C. Join AC. Then BAC 
will be 50°, as required. 

A greater angle than 60° cannot 
be taken from the sector with one measurement ; if the angle to 
be measured is more than 60°, successive measurements must be 
taken. 

144. On an arc f inch in radius, to construct an angle of 125°. 

Make the transverse distance between the points 60 and 60 
I inch. Let AB (Fig. 201) represent this distance. Describe 
an arc with AB as radius. Take 
the distance between the points 
50 and 50 from the sector, and 
set it off on the arc from B to C. 
Also take the distance from 40 
to 40, and set it off from C to 
D. Then take the distance 
between 35 and 35, and set it 
off from D to E. Join EA. Then the angle BAE will be 125°. 
5o° + 4o° + 35°-i25°. 

145. To construct an angle of 3^ on the same arc. 
With the sector open at the same angle as before, take the 
transverse distance between the points 47 and 47, and set it off 
on the arc from B to H. Join HA and CA. Then HAC will 
be 3° as required. 50° -47° = 3° 




loo GEOMETRICAL DRAWING AND DESIGN. 



The Line of Polygons. 

This pair of lines is used for dividing a circle into any number 
of equal parts between four and twelve, by joining which the 
regular polygons are formed. The transverse distance between 
the points 6 and 6 is always used for the radius of the circle to 
be divided ; because the radius of a circle containing a six-sided 
figure, i.e. a hexagon, is always equal to one side of the figure. 

Open the sector till the transverse distance between 6 and 6 
is equal to the radius of the circle ; then the distance between 
the points 4 and 4 will divide the circle into four equal parts, 
the distance between 5 and 5 into five equal parts, and so on 
up to twelve. 

If it be required to construct a polygon on a given straight 
line, open the sector till the transverse distance between the 
numbers answering to the number of sides of the required 
polygon shall equal the extent of the given line, then the distance 
between the points 6 and 6 will give the radius of the circle to 
be divided by the given line into the required number of equal 
parts. 

146. On a given line 1 inch in length, to construct a heptagon. 

Open the sector till the transverse distance between the 
points 7 and 7 shall equal i inch ; the distance between the 
points 6 and 6 will then give the radius of a circle, to which 
the given line will form seven equal chords. 



EXERCISES. 

1. On a line 4" long, draw a semicircle, and upon it set out the 
primary divisions of a protractor, by construction alone. 

2. Construct a scale of chords from the protractor set out in the 
preceding question. 

3. Make a scale of chords of 2" radius, to read to lO° up to 90°. 
The scale must be finished and figured. At the ends of a line 2^" long 
construct, from the scale, angles of 20° and 70° respectively. (April, '99. ) 



CHAPTER X. 



THE CONSTRUCTION OF SIMILAR FIGURES. 
PRINCIPLES OF SIMILITUDE. 

Similar Figures. 

Similar figures have their angles equal and their corresponding 
sides proportional. 

All regular figures — such as equilateral triangles, squares, 
and regular polygons — are similar. Other quadrilateral figures — 
triangles and irregular polygons — can be constructed similar to 
given ones by making their angles equal. 

147. To construct within a given triangle ABC, and equidistant 

from the sides of it, a 

similar triangle, the 

base of which is equal 

to the given line D, 

Fig. 202. 
Bisect the angles BAC 
and ACB by lines meet- 
ing at the centre E (Prob. 
12). Join EB. On the 
line AB set off AF equal to 
the given line D. From 
F draw a line parallel to 
AE till it cuts EB at G 
(Prob. 3). From G draw 
a line parallel to BC till it 
cuts EC at H. From H 

draw a line parallel to AC till it cuts EA at K. 
KGH will be the similar triangle required. 




Fig. 202. 



Join KG. 



102 



GEOMETRICAL DRAWING AND DESIGN. 



148. To construct about a given triangle ABC, and equidistant 

from its sides, a similar triangle, the base of which is equal 

to a given line L. Fig. 202. 

Set off on the base AB produced, AN equal to the given line 

L. From N draw a line parallel to EA (Prob. 3) till it meets 

EB produced at O. From O draw a line parallel to AB till it 

meets EA produced at M. From M draw a line parallel to AC 

till it meets EC produced at P. Join PO. Then MOP will be 

the similar triangle required. 



To construct within a given square ABDC, and equidistant 




150. 



from its sides, a square, one 
side of which is equal to the 
given line E. 

Draw the diagonals AD and 
CB. From A set off AF along 
AB equal to the given line E. 
From F, parallel to AD, draw a 
line till it meets CB at G (Prob. 
3). With M as centre, and 
radius MG, set off the points H, 
K, and L, and join GH, HK, 
KL,andLG. Then HGLKwill 
be the square required. 



Fig. 203. 



To construct a triangle similar to a given triangle CDE, and 
having its perimeter equal to a given straight line AB. 

On the given line AB construct 
a triangle ABF similar to the 
given triangle CDE, by making 
the angles at A and B equal to 
the angles at C and D respec- 
tively. Bisect the angles at A 
and B by lines meeting at G. 
From G draw a line parallel to 
FB till it meets AB at L ; and 
also a line parallel to AF till it 
meets AB at H. Then HLG 
will be the triangle required. 




CONSTRUCTIOX OF SIMILAR FIGURES. 



103 



Principles of Similitude. 

Draw a rectangle ABDC, and join each angle to any 
E. Bisect EB in G, EA in F, ED in K, and EC in H. 
FG, GK, KH, and 
HF, then FGKH 



point 
Join 




will 
be a rectangle with 
sides one -half the 
length of the rectangle 
ABCD. If we draw 
the diagonals BC and 
GH, we shall find 
they are parallel to 

each other, and that i"ig-205. 

BC : GH as 2:1. If we take any point L- in BD and join 
it to point E, LE will intersect GK in M, and will divide GK 
in the same proportion as BD is divided. 

On the principle of this problem, we can draw a figure similar 
to a given figure, and having any proportion desired, e.g. — 

If we wish to draw a rectangle having sides equal to one- 
third of a given rectangle, we should trisect the lines drawn 
from the angles of the given rectangle to E. The point E is 
called the centre of similitude of the two figures ABDC and 
FGKH, which in this instance are said to be in direct similitude. 
The following problem shows the principle of inverse similitude. 

151. To draw a trapezium similar to a given trapezium ABDC, 
with sides two -thirds the length of those of the given 
trapezium, In INVERSE SIMILITUDE. 
Take any point E in 
a convenient position. 
Produce a line from A 
through E to K, mak- 
ing EK equal to two- 
thirds of AE (Prob. 
117). Proceed in the 
same manner with each 
of the points B, C, and 
D, which will give the points H, G, and F. Join KH, HF, FG, 
and GK. Then FGHK will be the trapezium required. 




104 



GEOMETRICAL DRAWING AND DESIGN. 



152. To draw an irregular polygon similar to a given polygon, 
but with sides two-thirds the length of those of the given 
polygon ABCDEFG. 
It is not necessary that the centre of simihtude should be 

taken outside the figure ; if more 
convenient, we can use one of 
the angles of the figure, e.g. let 
A be the centre of similitude. 
Draw lines from all the angles 
of the polygon to A. Make AK 
two-thirds of AD (Prob. 117), 
and divide all the other lines in 
the same proportion, which will 
give the points L, M, N, O, H, 
andK. Join NM, ML, LK, KH, 
and HO, which will give the polygon required. 

Note. — The figures are similar, and the corresponding sides 




Fig. 207.' 



are parallel to each other. 



153. To draw an irregular pentagon similar to a given pentagon 
ABCDE, but with sides one-half the length of those of 

the given pentagon, 
without using any 
centre of similitude. 
Draw FG in any 
convenient position 
parallel to AB, and 
half its length (Prob. 
4). From G draw 
GH parallel to BC 
and half its length. 
Proceed in the same 
way with the remaining sides, which will give the pentagon 
required. 





154. 



To draw a curve or pattern similar to a given figure, 
but to two-thirds the scale. 



Enclose the given figure in a convenient rectangular figure 
ABDC, and divide the sides of the rectangle into equal parts 



CONSTRUCTION OF SIMILAR FIGURES. 



105 



(Prob. 9). Join these divisions, which will divide the rectangle 
into a number of equal squares or rectangles. Draw another 



G 




> 


^ 




H 






/ 


/ 


"^ 










, 


/ 


^ 






,\ 












— 


^ 
















E 










F 



r 


> 


-^ 






[ 


/ 


7 


-^ 




\ 




/ 


/ 


^ 


\ 


A 


/ 


/"^ 




^:^V(^ 








^^ 


w 





Fig. 209. 



Fig. 210. 



rectangle EFGH with sides two-thirds the length of the 
rectangle enclosing the given figure, and divide it in a similar 
manner. 

Draw the curves to intersect these smaller divisions in the 
same places as the larger divisions are intersected by the given 
figure. 

Note. — This method is used for enlarging or reducing maps 
or drawings to any scale. 



EXERCISES. 

1. Draw a regular hexagon on a side of i"; construct a similar figure 
on a side of ^", using one angle as the centre of similitude. 

2. Draw a rectangle with sides of 2 7" and 1.5" ; construct a similar 
figure by inverse similitude, with sides in the proportion to those given 
as 3 : 5- 

3. Draw a regular pentagon in a circle of 2§" diameter ; construct a 
similar figure by direct similitude, with sides in the proportion to those 
of first pentagon as 4 : 7. 

4. Make an irregular pentagon ABODE from the following data : 

Sides: AB=:2i", BC=ii", CD = 2r, DE=i". 
Angles: ABC =105°, ABE = 30°, BAE=io5°. 
Then make a si7Jiilar figure in which the side corresponding to AB is 
If" long. (June, '98.) 



io6 



GEOMETRICAL DRAWING AND DESIGN. 



5. Construct a rectangle hiavinq: one of ils sides if", and its diagonals 
2" long. Make a similar rectangle having its shorter sides i^" long. 

(April, '96.) 

6. Construct a triangle, sides i^", i|", and 2^" long, and a similar 
triangle having its longest side 2^" long. Measure and write down the 
number of degrees in each of the angles. (Tune, '97.) 

7. Construct a figure (Fig. 211) similar to that given in the diagram, 



but having the side corresponding to AB 2" long. 



(May, '97.) 




Fig. 212. 




Fig. 213. 



8. Construct a figure similar to the given figure (Fig. 212), but having 
the distance corresponding to CD i" long. (June, '99.) 

■ 9. Make a figure similar to the given one (Fig. 213), but having the 
length corresponding to CD 2" long. (April, '00.) 



CHAPTER XL 



CONIC SECTIONS. 




A conic section is obtained by intersecting a cone by a plane. 
There are fiv^e different sections to a cone, viz. : 

1. A triangle, when the plane cuts the cone 
through its axis. 

2. A circle, when the plane cuts the cone 
parallel to its base, as at A, Fig. 214. 

3. An ellipse, when the plane cuts the cone 
obliquely, without intersecting the base, as at B. 

4. A parabola, when the plane cuts the cone 
parallel to one side, as at C. 

5. An hjrperbola, when the cone is cut by a plane that is per- 
pendicular to its base, i.e. parallel to its axis, as at D, or inclined 
to the axis at a less angle than the side of the cone. 

These curves can be drawn with the greatest accuracy and 
facility by the following arrangement. Cut a circular opening in 
a piece of thin card-board or stiff 
paper, and place it a short dis- 
tance from a lighted candle ; this 
will form a cone of light (Fig. 
215). If we place a plane, e.g. 
a piece of paper pinned to a 
drawing-board, so as to allow 
the hght coming through the 
circular aperture to fall upon it, 
we can, by placing it in the 
several positions, intersect this cone of light so as to form 
the required sections, which can then be traced. C is the 
candle, A is the circular aperture, and P the plane. 




GEOMETRICAL DRAWING AND DESIGN. 



In Fig-. 215 the plane is parallel to the aperture, so the section 
obtained is a circle. 




Fig. 216. 



Fig. 217. 



If the plane is placed obHquely to the aperture, as in Fig. 216, 
the section obtained is an ellipse. 

By placing the plane parallel to the side of the cone, as in 

Fig. 217, we get as sec- 
^^''' tion, the parabola. 

If we place the plane 
at right angles to the 
aperture, we obtain the 
hyperbola, Fig. 218. 

By adjusting the posi- 
tions of the candle, aper- 
ture, and plane, we can 
required condition, both as 




Fig. 2 



obtain a conic section to suit any 
to shape or size. 

A truncated cone or frustum is the part of the cone below any 
section as A or B, Fig. 214. 



The Ellipse. 

An ellipse has two unequal diameters or axes, which are 
at right angles to each other. The longer one is called 
the transverse diameter, and the shorter one the conjugate 
diameter. 

The transverse diameter is also called the major axis, as 
AB (Fig. 219), and the conjugate diameter the minor axis, 
as CD. 



CONIC SECTIONS. 



109 




Fig. 219. 



155. The two axes AB and 
CD being given, to con- 
struct an ellipse. 
Take a strip of paper and 
set ofif upon it the distance 
FH, equal to half the major 
axis and the distance FG, 
equal to half the minor axis. 
By keeping the point G on 
the major axis, and the 
point H on the minor axis, 
the point F will give a 
point in the ellipse. A 
succession of points can be 
found in this manner, 
through which draw a fair 
curve, which will be the 
required ellipse. 

156. To construct an ellipse, given an axis and two foci. 

An ellipse has two foci, as the points A and B, Fig, 220, 
and the sum of the radii from these two points is always 
equal. 

Let A and B represent 
two pins, and ABC a piece 
of thread. The point of a 
pencil is placed inside the 
thread at C and moved 
so as to keep the thread 
always tight ; the point 
will trace out an ellipse. 
As the length of the thread 
is constant, the sum of 
the two radii is constant 
also. 

The length of the major 
or minor axis given will 
determine the length of the 
thread. 




Fig. 220. 



no GEOMETRICAL DRAWING AND DESIGN. 

157. To construct an ellipse by means of intersecting lines, the 

transverse diameter AB and the conjugate diameter CD being 

given. 

Draw the lines AB and CD bisecting each other at right 

angles in the point E (Prob. 5). Draw HK and FG parallel to 

AB, through C and D, 



H 



W\ 


E ' ' /' 


V^4 


,2> /2. -'I 



B 



and HF and KG parallel 

to CD through A and B. 

(Prob. 3). Divide AH and 

BK into any number of 

equal parts, say four (Prob. 

9), and AE and EB into 

the same number. Join 

C with the three points in 

AH and BK, and produce 

lines from D through the 

three points in AE and 

EB. Where these lines 

intersect those drawn from C, points in one-half of the ellipse 

will be obtained. Find corresponding points for the other half 

in the same manner, and draw a fair curve through the points 

obtained, which will be the required ellipse. 



D 

Figr. 221. 



158. To find the normal and tangent to a given ellipse ABCD, at a 
given point P. 

With C as centre, and 
radius equal to EA, draw 
the arc FH, which will 
give the two foci in F and 
H. Join the given point 
P with F and H, and 
bisect the angle FPH by 
the line PK (Prob. 12). 
PK is the normal. Draw 
the line NO through P, 
perpendicular to PK. 
This is the tangent re- 
quired. 




CONIC SECTIONS. 



Ill 



159. To complete an ellipse from an elliptical curve. 
Let AB be the given curve. Draw any two sets of parallel 

Join the points of bisection 




v^F 



chords and bisect them (Prob 

in each set by lines meeting 

in C. Produce one of these 

lines till it meets the given 

curve in D. With C as y^VA \ ^M 

centre, and CD as radius, set 

off on the given curve the 

point A. Join AD. Through 

C draw a line HK parallel 

to AD (Prob. 3), also the ^ 

line CL perpendicular to AD 

(Prob. 7). Produce CL to 

M, making CM equal to CL. 

Also make CK equal to CH. 

Then LM will be the transverse and HK the conjugate 

diameters. The ellipse can then be completed by any of the 

constructions already described. From A and D lines are drawn 

parallel to LM ; with C as centre, and radius CD, set off E 

and F, these will be two more points in the ellipse. 

160. To draw an ellipse to pass through three given points A, B, C. 

Join AC and bisect it in D. Join BD. From A and C draw 
the lines AE and CF parallel to BD. Through B draw the line 
EF parallel to AC. E B 

Produce BD to H, 
and make D H equal 
toBD. Divide AD 
and DC and also 
AE and CF into a 
number of equal 
parts, say four. Join 
the divisions on AE 
andCFtoB. From 
H, through the divisions on AC, draw lines till they meet the 
corresponding lines drawn to B. Draw a fair curve through 
these points, which will give half of the required ellipse. Proceed 
in the same way with the other half. 




112 GEOMETRICAL DRAWING AND DESIGN. 

161. To determine points for drawing a parabola, the focus A 
and the directrix BC being given. 

Draw the line EAD perpendicular to the directrix BC (Prob. y), 
which will give the axis. Bisect AD in F (Prob. i), which will 
be the vertex of the curve. Take any points a, d, c, d, and e in 
the axis, and draw perpendiculars through them. From A as 
centre, mark off on the perpendiculars, arcs with radii equal to 
aD, dD, cD, dD, and eD, cutting the perpendiculars in a'^ b\ ^, 
d\ and e' . These are the points required for the lower half of 
the parabola. The points above the axis are found in the same 
manner. 



B 






V 


A 




^ 


K/^ 


■^ 


f 

/' 


1 


\ 


-E 




c 


A 

\ 


-b 


C 

a' 


d L 


e 



Fig. 225. 



162. To draw a tangent to a parabola at a given point H. 
Join AH. From A set off AK on the axis produced equal 
to AH. Join KH, which will be the required tangent. This 
could also be found by drawing a Hne from H parallel to the 
axis till it meets the directrix in B, and then bisecting the angle 
AHB by the line KH (Prob. 12), which is the tangent. If from 
H we draw the line HL perpendicular to the tangent, it will be 
the normal. 



CONIC SECTIONS. 



113 



163. To draw a parabola, an abscissa AB and an ordinate BC 
being given. 

Complete the rectangle ABCD. Divide BC into any number 
of equal parts, say six 
(Prob. 9), and CD into 
the same number. From 
each division in BC draw 
lines parallel to CD (Prob. 
3), and from each of 
the divisions in CD draw 
lines to the vertex A. 
Where these lines of 
corresponding numbers 
intersect, e.g. where i 
intersects with i', 2 with 
2', etc., are points in the 
parabola. Find corre- 
sponding points on the opposite side of the axis, and draw a 
fair curve through them. 




164. To draw an hyperbola, the diameter AB, an ordinate CD, 
and an abscissa BD being given. 

Draw BE parallel to CD (Prob. 3), and complete the rectangle. 
Produce BD, and make 
AB equal to the given 
diameter. Divide CD 
and CE into any number 
01 equal parts, say four 
(Prob. 9), «, ^, c. The 
divisions on CD join to 
A, and those on CE to B. c' 
The intersection of the u 
corresponding lines, e.g. , 
where a intersects c^', b b\ ^ 
and c c\ are points in the 
hyperbola required. Find 
corresponding points for 

the other half, and draw a fair curve through them. 

H 




114 



GEOMETRICAL DRAWING AND DESIGN. 



A form of hyperbola frequently used is the rectangular hyper- 
bola. Let AB and AC 
represent two axes, and 
E the vertex of the curve. 
Complete the rectangle 
ABDC. Take any point 
H in CD and join it to 
A. Let fall a perpendi- 
cular from E till it meets 
HA in O (Prob. 7). 
From O draw OK parallel 
to AB till it meets a line 
from H parallel to AC in the point K (Prob. 3). This will be 
one point in the curve, and others may be found by taking fresh 
points on CD and treating them in a similar manner. 

One peculiar property of this figure is that, if we take any 
point in the curve and draw lines from it perpendicular to the 
lines AB and AC, — for example, KN and KM, — the rectangle 
contained by the two lines is always equal, i.e. KN x KM would 
be the same for any point in the curve. 




165. A mechanical method of drawing- a parabola or hyperbola. 
Let AB represent the edge of a drawing-board and CD the 

edge of a tee-square. Take 
a piece of string equal in 
length to CD, fix one end at 
D and the other at E, which 
is the focus of the curve. If 
a pencil be held against the 
string so as to keep it tight 
against the tee -square when 
the tee-square is moved up- 
wards, the pencil will trace 
half a parabola. AB is the 
directrix, and K the vertex of 
Fig. 229. the curve. Compare this 

method with the construction of Prob. 161. 

If the angle DCA were an acute or obtuse angle instead of a 

right angle, the pencil would trace an hyperbola. 




CONIC SECTIONS. 



"5 



166. To draw an oval by arcs of circles, its transverse diameter AB 
and its conjug-ate CD being given. 

Set off on AB the distance AE equal to half the conjugate 
diameter. Through E draw the line FG perpendicular to 

A 



AB (Prob. 7). With E 
as centre and EA as 
radius, draw the semi- 
circle CAD. From C 
and D set off CF and 
DG equal to EA. From 
B set off BH equal to 
half of EA. Join FH 
and GH. With F and 
G as centres, and FD, 
GC as radius, draw the 
arcs DL and CK. With 
H as centre, and H K as 
radius, draw the arc 



/ 

c/ 


^ 


E \ 


V i 




1 


-1 


> 


/ ' 




\N 


X / 




\ ^ 


^ / 






H^' / 




\ X 






\ ^ 


N X 




*K^ 


j/l 



D C, 



B 

Fig. 230. 



KBL, which wili complete the oval required. 

The conic sections are of frequent occurrence both in science 
and art : the heaAcnly bodies trace them in their courses ; they 
are used by engineers where A ^ 

great strength is required, such 
as the construction of bridges ; 
and they form the contour of 
mouldings, etc. Those subtle w 
curves that we admire in the 
outline of Japanesehandscreens 
and vases are often parabolas. 

Fig. 231 is an illustration 
showing how these curves are 
applied to art forms. 

To draw the curve CB. 
Draw the lines AB and AC 
at right angles to each other 
(Probi* 5). Di-^id^^ch into 
the same number of equal jjarl#^.^, 
(Prob. 9), and join them. Pro- 




Fig. 231, 



ceed in the same manner with the other curve** 



ii6 GEOMETRICAL DRAWING AND DESIGN. 



Cycloidal Curves. 

If a circle is rolled along a line in the same plane, a point in 
the circle will describe a curve of a class called cycloidal. 

The line along which the circle rolls is called a director, and 
the point itself is called the generator. 

The curve is called a Cycloid when the generator point is in 
the circumference of the rolling circle and the director is a 
straight line ; but a Trochoid when the point is not in the circum- 
ference of the circle. 

When the director is not a straight line, but the outside of 
another circle, and the generator is in the circumference of the 
rolling circle, the curve desc^bed is called an Epicycloid ; but 
when the point, or generator, is not in the circumference of the 
rolling circle, it is called an Epitrochoid. 

If the director is the inside of a circle and the generator a 
point in the circumference of the rolling circle, the curve is 
called a Hypocycloid ; but if the generator is not in the circum- 
ference of the rolling circle, it is called a Hypotrochoid. 

The Epicycloid and Hypocycloid are the true curves for the teeth 
of gearing. The director is the pitch circle of each wheel : and 
if the rolling circle be the same for the whole set, they will gear 
into one another. 

In constructing a cycloid it is necessary to make a line equal 
in length to the arc of a semicircle. The exact relation betwee" 
the diameter and circumference of a circle cannot be express 
in numbers ; but the following problem will enable us to arri- 
at an approximation, correct to six places of decimals. 

167. To draw a line equal to tlie length of a semicircle. 

Let AC represent the radius. Draw the semicircle ABD. 
Produce AC to D, and draw BC perpendicular to it. From A 
and D draw tangents parallel to BC, and through B drav/ a 
tangent parallel to AD. From B set off ;S£>»e(ifiial to the jadius, 
and draw the line BF through E« Pj-oduce the tangent through 
A to H, and makej^H equal to AD. Join HF, which will be 
the line require^ 



CONIC SECTIONS. 



If we take AC to represent a length equal to the 
of a circle, then HF will equal the circumference, 



to the dimneter 




F D 

Fig. 232. 

We are also enabled to find the length of an arc by this 
means ; e.g: the arc to the chord formed by one side of a pentagon, 

' irfH IS- eq3.l.tn j^he circumference, then 
arc required. 



FH 



-theleno-fh-(;,;\,,e 



168. To draw a cycloid. 

Let AB be the director, and / the generator or point in the 

rolling circle AM/). Draw Ak equal in length to half the 

circumference of the circle AM/, and divide it into any number 

^ "" equal divisions (Prob. 10), say six, at d, e,/, j^, and /t. Divide 

the semicircle into the same number of equal divisions (Prob. 

14), and draw lines from each division parallel to the director 

AB. Draw the line CK from the centre of the circle parallel to 

AB. Draw lines perpendicular to AB at the points d, e,/,g; /i, 

and k till they meet the line CK in the points D, E, F, G, H, 

and K. With each of the points D, E, F, G, H, and K as 

mtres, and a radius cqunl to Cp, draw arcs cutting the parallel 

^^»f^l^m^tltei the dnisions ii^'tb^ fliiftlicircle in the points 

K^ ^f f^,f3lV »hid T. This wilT give points in half the cycloid. 

ndf:<^ 40rt-^s(pWiciwig' rtcjfnts for the remaining half, and draw 

feU'CtiWfe' through tlve p«hir ;v^^liH give the cycloid 

furtTe'd. 



ii8 



GEOMETRICAL DRAWING AND DESIGN. 



To determine the tangent and normal to the curve at any 
point /:— Draw the Hne //' parallel to AB till it meets the 




Fig. 233. 

generating circle at /'. Join /'A. Through / draw the line 
WW parallel to /'A. This will be the normal to the curve. The 
tangent rs is at right angles to this line. 



169. To draw an epicycloid. 



Note. — The length of the di 



a complete curve is 



to the whole circle as the radius of the rolling circle is to the 
radius of the director; e.g. if radius of rolling circle == i inch, and 
that of director = 6 inches ; then the director = ^ of the circle. 

Let AB be the director, which is a part of a circle, and/ the 
generator. Take Ak equal in length to half the rolling circle, 
AM/, and divide it into any number of equal divisions, say six, 
at ^, ^, /, g., and //. Divide the semicircle into the same number ; 
of equal divisions, and draw lines from these points, as well as 
from the centre of the circle, concentric with the arc /:.B. 
From the centre of the circle that contains the arc AB draw 
lines through the points d, e,f,g., h, and k till they meet the arc 
drawn from the centre of the rolling circle. With D, E, F, G, 
H, and K as centres, and a radius equal to Cp, draw arcs til) 
they meet the concentric arcs drawn from the divisions of th'^e 
semicircle in the 
sponding points^ 
through all the 

At any poi^i^f -^^WF"""^BHHm|^^^^^K c .^^^' 
proceed ^,gJpllow^. ^R'^ ^^^ arc tt concent^^MPp^'--AB» 




t:> iy(X 



CONIC SECTIONS. 



119 



till it meets the generating circle in /'. Join /'A. With / as 
centre, and radius equal to /'A, draw an arc intersecting AB at 




Fig. 234. 



w'. Join tu>\ and produce it to w. 
tangent rs is at right angles to it. 



This is the normal. The 



170. To draw a hypocycloid. 

Let AB be the director, which is the arc of a circle, and p 
the generator, which is a point in the circumference of the 
roUing circle M. Make A/j equal in length to half of the circle 
M, and divide it into any number of equal parts, say six, at d^ 
^■> /■> .^1 arid h. Divide the semicircle into the same number of 
equal parts, and from, the centre of the circle containing the arc 
AB draw concentric arcs from these points, ^s well as from the 



centr 



. e C. Draw lines from the points d, e^f, g, h^ and k towards 

the centre of the circle co-^maining the arc AB till they meet the 

\rQ from the centre C in the points D, E, F, G, H, and K. With 

- jt oe^ noints as centred, and a radius equal to C/, draw 

res till they meet the concentric arcs drawn from the divisions 

■ the semicircle in the points N, O, p-," O, R, and T. Find 

: corresponding points for the othef half, and draw a fair 

•/e through all the points, which will give the hypocycloid 

-juired. 



I20 GEOMETRICAL DRAWING AND DESIGN. 

The tangent and normal at any point / are thus o i. ed. 
Draw the arc W concentric to the arc AB till it meets ihe 
generating circle at /'. Join /'A. With / as centre, and radius 




Fig. 235. 

equal to /'A, set off on AB the point w'. Join fw', and produce 
it to 7U. This is the normal. The tangent rs is at right 
ano-les to it. 



171. To construct a continuous curve, by a combination of arcs of 
different radii, through a number of given points A, B, C, D, E, 
F, G, and H. 

Join the points AC, CD, DE, etc. Find the centre K of the 

circle containing the arc 
ABC (Prob. 35). Join CK. 
Bisect the line CD at right 
angles, and produce the 
bisecting perpendicular till it 
meets CK produced in L. 
Join DL. Bisect the line-''^]g 
'^i^'A produce The oisecting 
\ierpendicular till it meets 
DL produced in '^l. Find 
the remaining points N, O, 
and P in the same manner. 
'^' '^ ■ • The points K, L, M, N, O, 

and P are the centresr^of the circles containing the arcs 
necessary for joining the given points. ; 




CONIC SECTIONS. 



njiijdf^Q construct an Arcliimedean spiral of one revolution. 
Draw a circle and divide it by radii into any number of equal 
parts, say twelve (Prob. 9) a^ b^ 
c, d^ etc. Divide the radius o 
into a corresponding number 
of equal parts i, 2, 3, 4, etc. 
(Prob. 9). From the centre of 
the circle, with radius i, draw 
an arc till it meets the radius a 
in A, and from 2 till it meets 
the radius b in B, and so on till 
the whole twelve are completed. 
Draw a fair curve through these 
points, A, B, C, D, etc., which 
will give the spiral required and 
proceed in exactly the same manner for further revolutions. 




173. To draw the logarithmic spiral.. 

The logarithmic spiral was discovered by Descartes, 
also called the equiangular 
spiral, because the angle the 
curve makes with the radius 
vector is constant. The curve 
also bears a constant proportion 
to the length of the radius 
vector 

Take any line AC for the 
radius vector, and bisect it in D 
(Fig. 238). With D as centre, 
and radius DA, draw the semi- 
circle ABC. From the points 
A and C draw any two lines AB 
and CB cutting the semicircle. 
Then ABC is a right-angled 
triangle. 

Bisect the Hne BC in E. 
With E as centre, and EB as 
"''^ius, draw the semicircle 



It is 




GEOMETRICAL DRAWING AND DESIGN. 



BFC. Make the angle BCF equal to the angle ACB, and pro- 
duce the line till it meets the semicircle in F. Join BF. The 
triangle BFC is then similar to the triangle ABC. By repeating 
this construction we obtain a succession of similar triangles 
radiating from a common centre C, and all forming equal 
angles at this point. The exterior points of these triangles, viz. 
A, B, F, M, N, O, etc., are points in the required spiral. 

As each triangle with its curve forms a similar figure, it is 
evident that the curve must form a constant angle with its 
radius vector,' i.e. the line radiating from C, and the portion of 
the curve accompanying each triangle, must also bear a constant 
proportion to the length of its radms vector. 

If we bisect the angle ABF by the dotted line HB, this line 
will be the normal to the curve ; and the line KL, being drawn 
at right angles to HB, is the tangent to the spiral. 

As all the angles at C are equal, the spiral could be con- 
structed with greater facility by 
first drawing a circle and divid- 
ing it into an equal number of 
parts by radii, as shown in 
Fig. 239. 

Let AC be the radius vector. 
Bisect it in D. With D as 
centre, and radius DA, draw an 
arc cutting the next radius CB 
in B. Proceed in the same 
manner with each radius in suc- 
cession, which will determine 
the points H, K, L, M, N, O, 
P, etc. Draw a fair curve 
through these points, and we shall obtain a logarithmic spiral. 

The greater the number of radii used in the construction, 
the larger will be the angle BAC ; but the angle ABC will 
always be a right angle, as will be seen by the construction in 
Fig. 238. 




Fig. 239. 



174. To draw a spiral adapted for the Ionic volute by means of arcs. 

Divide the given height AB into eight equal parts (Prob. 9). 

Bisect the fourth part in the point C (Prob. i), and from it 



CONIC SECTIONS. 



[23 



draw a line perpendicular to AB (Prob. 7). 
equal in length to four of the divisions of AB 
the eye of the volute D. This 
is shown to a larger scale at E. 
With D as centre, draw a circle 
with a radius equal to C4. In- 
scribe a square in this circle 
(Prob. 56), and bisect each of 
its sides in the points i, 2, 3, 
and 4 (Prob. i). Join these 
points, and draw diagonals. 
Divide each semi-diagonal into 
three equal parts and join them 
(Prob. 9), thus making three 
complete squares parallel to 
each other. The corner of 
each of these squares in succes- 
sion will be the centre of one 
of the arcs, commencing at i, with 
dotted hnes and arrow- heads. 



Make this line 
which v/ill sive 




L as radius, as shown by 



EXERCISES. 

1. Construct an ellipse: major axis 3.75", minor axis 2.25". Select 
any point in the curve, and draw a tangent to it. 

2. Construct an ellipse ; the foci to be 2^" apart, and the transverse 
diameter 3f". 

3. Draw a rectangle 3.25" x 2.3", and inscribe an ellipse within it. 

4. Draw a parallelogram 3I" x 2|", two of its angles to be 60° ; 
inscribe an ellipse within it. 

5. Construct an ellipse by means of a paper trammel ; the transverse 
diameter being 4I", and the conjugate diameter 3" (Prob. 155). 

6. Draw the two diameters of an ellipse each 3" long, and at an 
angle of 45° with each other ; complete the ellipse. 

7. Make a tracing of the ellipse given in question 3, and find the 
diameters, foci, tangent, and normal. 

8. With an abscissa 3" long and an ordinate 2" long, construct a 
parabola. 

9. With a diameter 1.4", an ordinate 1.8", and an abscissa 1.4", 
construct a hyperbola. 



124 GEOMETRICAL DRAWING AND DESIGN. 

10. Draw a rec'angle 3" x 2", and let two adjacent sides represent 
the axes of a rectangular hyperbola ; measure off along one of its longer 
edges ^", and let this point represent the vertex of the curve ; complete 
the hyperbola. 

11. Draw a line 4" long to represent an abscissa of a parabola ; at one 
end draw a line 3" long, at right angles to it, to represent the directrix ; 
from the directrix, along the abscissa, set off i" to mark the focus ; 
complete the parabola. At any point in the curve, draw a tangent and 
normal to it. 

12. Draw geometrically an ellipse, a parabola, a hyperbola, each 
2 inches long, and a cone 2 inches high, and write the name to each. 
Show the following five sections on the cone — a horizontal section, a 
vertical one, not through the apex, and one through the apex, one 
parallel to one side, and one cutting both inclined sides. Name figure 
each section makes. (May, '96.) 

13. The foci of an ellipse are 2|" apart and its major axis is 3^" long. 
Describe half the curve. (April, '96.) 

14. The foci of an ellipse are 2|" apart, and its minor axis is 2" long. 
Draw the curve, and draw also a tangent from a point on the curve \" 
from one of the foci. (June, '00.) 

15. Draw the curve (Fig. 241) every point of which is at equal 
distances from the line PQ and the point F. The curve, which is a 
parabola, need not be shown below the line ST. 

(April, '96.) 



Dimensions to be trebled. Fig. 242. 

Fig. 241. 

16. Draw a straight line AB 3" long. Bisect AB in F. At F draw 
FV f" long at right angles to AB. F is the focus, and V the vertex of 
a parabola, A and B being points on the curve. Draw the curve from 
A to B, showing the construction for at least 4 points. (June, '00. ) 

17. Two conjugate diameters of an ellipse are 3^" and 2|" long 
respectively, and cross one another at an angle of 60°. Draw the curve. 

(April, '99.) 

18. An arch in the form of a semi-ellipse is 6' wide and 2' high. 
Describe the curve, and draw two lines perpendicular to it from two 
points on the curve, each 2' from the top point of the arch. 

Scale (which need not be drawn) ^" to i'. (April, '98.) 



CONIC SECTIONS. 125 

19. Draw the arch opening shown (Fig. 242), using the figured dimen- 
sions. The curve is a semi-elUpse, of which I'Q is a diameter, and RS 
is half its conjugate character. (June, '99.) 

20. Describe the spiral of Archimedes of three revolutions, whose 
radius is 2 inches. (April, '98.) 

21. Within a circle of 2" radius describe a ' spiral of Archimedes ' of 
one revolution. (June, '99.) 

22. Sketch the three sorts of spiral, and explain how each is gene- 
rated, and illustrate each sort by shells, plants, or animals. (May, '96.) 



PART IL 



SOLID GEOMETRY. 



CHAPTER XII. 



INTRODUCTION. 



In the preceding subject, Plane Geometry, we have been 
restricted to figures having length and breadth only, but Solid 
Geometry treats of figures that have thickness in addition to 
length and breadth. 

The objects taken to illustrate the principle of this subject 
are described under the head of Definitions, Solids (page lo). 

By means of Practical Solid Geometry we are enabled to 
represent on a plane — such as a sheet of paper — solid objects in 
various positions, with their relative proportions, to a given scale. 

Let us take some 
familiar object, a 
dressing-case for in- 
stance, ABDC, Fig. 
243, and having pro- 
cured a stiff piece 
of drawing-paper 
HKLM, fold it in a 
line at X, parallel to 
one of its edges ; 
then open it at a right angle, so that HX will represent the 
edge of a vertical plane, and XL the edge of a horizontal plane ; 
the line at X, where the two planes intersect, is called the line 
of intersection, intersecting line, or ground line ; it shows where 




INTRODUCTION. 



127 



the two planes intersect each other, and is generally expressed 
by the letters X and Y, one at each end. 

Having placed the dressing-case on the horizontal plane, 
with its back parallel to the vertical plane, let us take a pencil 
and trace its position on the jj g- 

horizontal plane by drawing a 
line along its lower edges ; also 
its shape on the vertical plane. 
This can be done by placing 
the eye directly opposite each 
of its front corners in succes- 
sion and marking their apparent 
position on the vertical plane, 
and joining them. Having done 
this, we will remove the dress- 
ing-case and spread the paper 
out flat upon a table : this is 
shown in Fig. 244. We have now two distinct views of the 
object. The lower one is called a PLAN, and represents the 
space covered by the object on the horizontal plane, or a view 
of the dressing-case seen from above. The upper view shows 
the space covered on the vertical plane, and is called an 
ELEVATION : it represents the front view of the object. 

In Solid Geometry all objects are represented as they would 
appear traced or projected on these two planes at right angles 
to each other : they are b 
called co-ordinate planes. 








/ 


J' 








B 


X 










c 




d 









a 




b 


L 





M 



Fig. 244. 




It is not necessary that 
the object should be 
parallel to them, as in 
Fig. 243 : we can arrange 
it in any position, making 
any possible angle with X 
either plane, but the hne 
connecting the point on 
the object with its respec- 
tive plane must always be perpendicular to that plane. We 
shall understand this better if we refer to Fig. 245, in which we 
will imagine the dressing-case suspended in mid-air, with its back 



128 



GEOMETRICAL DRAWING AND DESIGN. 



still parallel to the vertical plane, but its under side inclined 
to the horizontal plane. We will now trace it on each plane as 
before, then by spreading the paper out flat we get a drawing 
as shown in Fig. 246. 

The student should compare Fig. 243 with Fig. 244, as well 
as Fig. 245 with Fig. 246, so as to thoroughly understand 
the relation between the co-ordinate planes. 

The lines Aa\ B^', Y.e (Fig. 246), are all perpendicular to the 
vertical plane ; and the lines A«, B/^, Cc, etc., are perpendicular 

to the horizontal plane. These 
lines are called projectors, and 
are here represented by dotted 
lines. The points in which 
these lines meet the two co- 
ordinate planes are called pro- 
jections : if they are on the 
vertical plane they are called 
vertical projections, and if on the 
horizontal plane horizontal pro- 
jections, of the different points ; 
e.g. a! i§ the vertical projection of 
point A, and a is its horizontal 
projection. The length of the 
horizontal projector shows the 
distance of the point from the vertical plane, and the length 
of the vertical projector its distance from the horizontal plane. 

This method of representing solid objects by projection on 
two planes is called orthographic projection, and is described 
more in detail in Chapter XIV. As the projectors are parallel 
to each other, it may also be called parallel projection. 

All through this subject the points of the object are dis- 
tinguished by capital letters, as A, B, etc., while their horizontal 
projections are represented as <2, b., etc., and their vertical pro 
jections as «', b\ etc. ; by this means we are enabled to dis- 
tinguish the plan from the elevation. v.P. will also be used to 
express the vertical plane, and h.p. the horizontal plane ; the 
letters XY will always stand for the ground line. 

The student should take particular notice that the lower points 
in the plan always represent the front points in the elevation. 




Fig. 246. 



INTRODUCTION 



129 



It is not necessary to have an object to trace ; if we know 
its dimensions, and its distances from the two planes, we can 
construct the plan and elevation as shown in Figs. 244 and 246. 

Figs. 243 and 245 are perspective views, and Figs. 244 and 246 
are geometrical drawings of the same object. If the latter 
were drawn to scale, we could find out the length, breadth, and 
thickness of the object from these drawings. 

Each perspective view is supposed to l3e taken from one 
fixed point, i.e. the eye ; and lines drawn from different parts of 
the object converge towards the eye considered as a point. 
These lines represent rays of light from the object, and are 
called visual rays : they form a cone, the vertex of which is the 
position of the eye ; consequently, Perspective is called conical, 
radial, or natural projection, because it represents objects as they 
appear in nature. It is impossible to see an object as it is 
represented by orthographic projection. 

We will now take four simple solids, viz. a cube, a rectangular 
solid, a pyramid, and a triangular prism, and show the different 
positions they can occupy with reference to the co-ordinate 
planes, i.e. the v.p. and H.P. 

Fig. 247 represents the four solids in what is called simple 
positions, i.e. parallel to both the v.p. and H.P. 



H 



'nx 









D 











Fig. 247. 

A is the plan of the cube and A' its elevation. 
B „ rectangular solid „ B' „ 

C „ pyramid „ C „ 

D ., triangular prism „ D' „ 



130 GEOMETRICAL DRAWING AND DESIGN. 




Fig. 248. 
Fig. 248 represents the same solids with their bases on the H.P. 
as before, but their sides are now inclined to the V.P. 






Fig. 249. 

Fig. 249 shows them with their fronts and backs, parallel to the 
V.P., as in Fig. 247, but with their bases inclined to the H.P. 




Fig. 250. 



INTRODUCTION. 



Fig. 250. — They are here represented inclined to both the v.p. 
and H.P., but they still have one set of edges parallel to the H.P. 

Fig. 251. — Here they are shown with every line inclined to both 
planes : instead of having one ecige resting on the H.P., as in 




Fig. 25 



Fig. 250, they are each poised on a corner. To distinguish this 
position from the one illustrated in Fig. 250, we will call it com- 
pound oblique; although Figs. 250 and 251 generally come under 
one head, as objects inclined to both planes^ 



CHAPTER XIII. 

SIMPLE SOLIDS IN GIVEN POSITIONS TO SCALE. 

Note, — Feet are indicated by one dash, and inche5i by two 
dashes : thus — 3' 6" represents 3 feet 6 inches. 

The student should draw the problems in Solid Geometry 
to a scale three times that of these figures. 



175. To project a quadrilateral prism 5" x 2^" x 2h" with one of its 
smaller faces on the H.P., parallel to the v.p., and |" from it. 
Scale } full size. Fig. 252 a. 

First draw the line XY ; then draw the plan abc'c;^ | inch 
below it. Draw perpendicular lines above XY, immediately 

over the points a and 6, 
^l>' 5" in height, which give 

the points a' and b'. Join 
a'b'. This is the elevation 
of the solid. 

176. To project the same 
solid with one of its 
longer faces resting on 
the H.P., parallel to 
the V.P., and l|" from 
it, to the same scale. 
Fig. 252 B. 

Draw the plan e/^^^, 
S"X2^", and if inches 
below XY. Draw perpendiculars above XY, 2^" high, and 
directly over the points e and /, which will give the points ^' 
and/'. Join <?' and/, which completes the elevation. 




Fig. 252. 



SIMPLE SOLIDS. 



^33 



The student should now project the four sohds illustrated 
in Fig. 247 in the positions there shown, but to the following 
dimensions and scale : 

A to have a base 4" x 4", to be 4" high and 2h" from v. p. 
B „ „ 8"x4" „ 2" „ 2^" „ v.P. 
C „ „ 4"X4" „ 8" „ 2i" „ v.P. 



D 



6"X4" „ 4" 
Scale J full size. 



v.P- 



177. To project a quadrilateral prism 10 " x 5" x 5", with one of its 
smaller faces on the h.p., at an angle of 45' with v.p., and 
one edge 3^" from v.P. Scale ^^ full size. Fig. 253 A, 

DrawXY. Take 
the point ^, 3^" 
below XY, and 
draw the square 
abed below this 
point with its sides 
at an angle of 45° 
with XY. This 
will be the plan. 
Erect perpendicu- 
lars 10" high above 
XY, and directly 
over the points a, 
b, and c. Join the 
tops of these per- 
pendiculars, which completes the elevation. 





I 


' 
















g- 


' t 




r 


X 








Y 










A 






',d 










a 




c g 


r" 


y 


f 






/ 


I- 253- 


h 


', 





178. To project the same solid lying on one of its longer faces on 
the H.P. with its longer edges forming an angle of 30° with 
the V.P., and one of its corners l|" from v.P., to the same 
scale. Fig. 253 P.. 

Draw the point /i, i|" below XY, and construct the plan £/(^/i 
at the required angle below this point. Erect perpendiculars 
5" high above XY, immediately over the points o-^ e, and /. 
Join the tops of these perpendiculars, which completes the 
elevation. 



134 



GEOMETRICAL DRAWING AND DESIGN. 



The student should now project the four soHds illustrated in 
Fig. 248, in the positions there shown, but to the following 
dimensions and scale : 
A to have a base 5" x 5", to be 5" high, with one side inclined at 

an angle of 30° with the V.P., and 2" from it. 
B to have a base 10" x 5", to be 25" high, with both sides inclined 

at an angle of 45° with the V.P., and 2" from it. 
C to have a base 5" x 5", to be 10" high, with one edge of the base 

making an angle of 60° with the V.P., and its nearest point 

2" from it. 
D to have a base 8" x 5", to be 5" high, with both sides of its base 

making an angle of 45° with the v.?., and the nearest point 

2" from it. 



179. To project a quadrilateral prism 7i"x3|"x3|", resting- on 
one of its shorter edges on the H.P., and with its longer edges 
parallel to the v. P., but inclined at an angle of 60° to the H.P. ; 
one of its faces to be 2^" from v.p. Scale J full size. Fig. 254 B. 




Draw XY. At point e' draw the elevation e'a'h'c' at the 
required angle. 2J" below XY draw the line df parallel to it. 



SIMPLE SOLIDS. 



m 



Let fall lines from a', b', and c\ at right angles to XY, and 
make da and /c each 3|" long. Join ac^ which completes the 
plan. 

The student should now project the four solids illustrated in 
Fig. 249 from the following conditions : 
A to be 2^" X 2^^" X 2^", with its base inclined at an angle of 45° 

to H.P. ; to be parallel to the v. p., and 2^" from it. 
B to be 5"x2V'x i^", with its base inclined at an angle of 30° 

to H.P. ; to be parallel to the v.p., and 2^" from it. 
C to be 2^"x2^"x 5", with its base incHned at an angle of 45° 

to H.P. ; to have the edge of its base parallel to the v.p., 

and 2h" from it. 
D to be 4" X 2|" X 2|", with its base inclined at an angle of 30° 

to H.P. ; to have its ends parallel to v.p., and if" from it. 
Scale ^ full size. 
Before proceeding with the next problem. Fig. 254 B, it will be 
necessary to understand thoroughly the angles which the solid 
forms with the co-ordinate planes. 

The longer edges are still inclined to the H.P. at an angle of 
60° ; but instead of being parallel to the v.p., as in Fig. 254 A, they 
axe in planes inclined to the 
V.P. at an angle of 45°. This 
does not mean that they 
form an angle of 45° with 
the V.P. Let us illustrate this 
with a model. 

Take a sheet of notepaper 
and draw a diagonal to each 
of its inside pages, as ad and 
dc, Fig. 255. Now stand it 
on a table against the 45° 
angle of a set-square. The 
two pages will then represent 
two planes at an angle of 45° 
with each other. Let the 
page a represent the V.P., 
and the line 6c one of the edges of the solid. The angle which 
dc makes with the page a will be considerably less than 45°. 




Fig. 25s 



136 GEOMETRICAL DRAWING AND DESIGN. 

180. To project a quadrilateral prism 7h" x 3|" x 3|", resting on 
one of its shorter edges on the H.P., with its longer edges 
inclined to the h.p. at an angle of 60°, and in vertical planes 
inclined to the v. p. at an angle of 45° ; one of its lower comers 
to be 1;^" from v.p. Scale | full size. Fig. 254 B. 

Find the position of point n, i^" below XY, and draw the 
lines nm and nk at an angle of 45° to XY. Make nm and n/c 
equal in length to ^ and da (Fig. 254 a). Draw the line k/ 
parallel to 7?m, and of the same length. Join /;;z. Make k^ 
and /o equal to ae and cd (Fig. 254 a). Draw^// and o^ perpen- 
dicular to /^/. This will complete the plan. As every point in 
the elevation is found in precisely the same way, it is only 
necessary to explain the projection of one point : <?', for example. 
Draw a perpendicular from o on plan till it meets a hori- 
zontal line drawn from d' (Fig. 254 a). This gives the position 
of point o'. 

The student should now project the four solids shown in Fig. 
250 from the following conditions : 
A to be 4"x4"x4", with one set of edges parallel to H.P. ; its 

other edges to be inclined to h.p. at an angle of 45°, and 

in vertical planes inclined to v.p at an angle of 30° ; its 

nearest corner to be 4" from v.p. 
B to be 8" X 4" X 2", with one set of edges parallel to H.P, ; its 

longest edges inclined at an angle of 30° to H.P., but which, 

with its shortest edges, are to be in vertical planes inclined 

to the V.P. at an angle of 30°. Its nearest corner" to be 4" 

from v.p. 
C to be 4"x4"x8" ; its base to be inclined to the H.P. at an 

angle of 45° ; its axis in a vertical plane incHned to V.P. 

at an angle of 30° ; and its nearest corner 4" from v. p. 
Note. — The axis is a line drawn from the vertex to the centre 
of the base ; as ai?, Fig. 250. 
D to be 6" X 4" X 4", to have one side inclined at an angle of 30° 

with H.P., and its ends in a vertical plane inclined to v.p. 

at an angle of 60°. The nearest corner to be 4" from V.P. 

Scale ^ full size. 

Note. — The heights in these elevations are obtained by first 
drawing side views- of the objects, as shown in Fig. 249. The 
connection is fully shown in Figs. 254 A and B. 



SIMPLE SOLIDS. 



137 



181. To project a quadrilateral prism 6j" x 3^" x 3|" resting on one 
corner on H.P., and its faces forming equal angles with it, with 
its longer edges inclined at an angle of 60° to H.P., and parallel 
to v.p. Its nearest edge to be Ij" from v. P. Scale ^ full size. 

Fig. 256 A. 

Draw XY, and ij" below it draw the line ;;/;;. In any con- 
venient position draw the line ca perpendicular to ;«;?, and 
from c draw c^ 3^" long, at an angle of 45°. From d as centre, 
and with radius dc, draw an arc cutting ca in a. Join L?a. Also 
draw M perpendicular to ca. This represents one-half of the 
actual shape of the base of the prism. 




Fig. 256. 



At any point c' on XY draw the line //' 6^" long at an angle 
of 60° to XY, and the line /// perpendicular to it. From /, 
along e7i', set off the distances e', ^', h\ equal to those of <?, <y, c. 
At each of these points draw lines parallel to e'f\ and equal to it. 
Join the tops of these lines. This completes the elevation. 
Draw lines from b and a parallel to i7in. Every point in the 
plan must come on one of these three lines. Drop a line from 
/' at right angles to XY till it meets the horizontal line from b \ 
this gives the point/ Every other point in the plan is found 
in the same manner. 



f38 



GEOMETRICAL DRAWING AND DESIGN. 



182. To project the same prism, with its longer edges stUl inclined 
at an angle of 60° with H.P., and its faces making equal angles 
with it ; taut instead of taeing parallel to v. p., as in Fig. 256 A, 
let them be in vertical planes, inclined at an angle of 45° with 
v.P. The nearest corner of prism to be Ij" from V.P. Scale J 
full size. Fig. 256 B. 

Note. — We can always imagine any line to be contained in a 
vertical plane, whether the object contains one or not. In Fig. 
254 B the line k'o' is contained in the vertical plane J^o'l'g' ; but 
in this instance s'w' (Fig. 256 b) is not contained in one, as the 
solid does not contain a vertical plane. 

At any point o ij" below XY, draw oq at an angle of 45° with 
XY. Make oq equal in length to nm (Fig. 256 a). The plan Fig. 
2 56 B is precisely similar in shape to the plan Fig. 2 56 A, but turned 
to make an angle of 45° with XY ; so if we take the line qo to 
represent the line nin^ we can complete the plan from Fig. 256 A. 

Every point in the elevation is found in the same way : erect 
a perpendicular upon point r till it meets a horizontal hne 
drawn from/' (Fig. 256 a) ; this gives the point r', and so on 
till the elevation is completed. 

183. To project a regular hexagonal prism 10" long, and with sides 

3^" wide, standing on its 
4r-4 taase on H.P., with one 

face parallel to v.P. and 
2^" from it. Scale jV 
full size. Fig. 257 A. 
Draw the Hne XY, and 
2|" below it draw the line 
ab. Complete the hexagon. 
Above XY draw perpendi- 
culars 10" long immediately 
above the points c^ e^f, d. 
Join the tops of these per- 
pendiculars, which com- 
pletes the elevation. 




184. To project the same prism lying on one face on the H.P., 
with its longer edges parallel to the v.P. ; its nearest edge to 
be if" from v.P. Scale ^V full size. Fig. 257 B. 
Draw the line /w, 10" long i|" below XY and parallel to it. 

Draw the lines Is and mp perpendicular to Im. Set off the 



SIMPLE SOLIDS. 



139 



.^ distances w, n^ <?, p on mp equal to the distances c\ /,/', d 
I (Fig. 257 a). Draw lines from «, o, and^, parallel to /w, till they 
meet the line /s. This completes the plan. 

Draw perpendiculars above XY immediately above the points 
s, pj and set off the distances u', s', r equal to the distances 
K, H, G (Fig. 257 a). Draw lines from the points / and r parallel 
to XY till they meet the perpendicular p'd. This completes the 
elevation. 



185. To project a regular hexagonal prism 7V' long, and with sides 
2^" wide, standing on its base on the H.P., with one face in- 
clined to the v.P. 
at an angle of 45". ^' c' ^' g* 
Its nearest edge to 
be l|" from v.P. 
Scale \ full size. 
Fig. 258 A. 

Draw the line XY, 
and if" below it mark 
the position of point a. 

From a draw the 
line ab^ i|" long, at an 
angle of 45° with XY. 
Complete the hexagon. 
Above XY draw per- 
pendiculars 7^" long 
immediately above the 
points b^ r, d^ e, and 
join the tops of these Hnes 




Fig. 258. 

This completes the elevation. 



186. To project the same prism, l3ring with one face on H.P. ; its 
longer edges to be inclined to the V.P. at an angle of 30° ; its 
nearest corner to be l\" from v.P. Scale |^ full size. Fig. 258 B, 

Fix the point/ i j" below XY. Draw the line f^, y\" long, 
at an angle of 30° with XY ; and from /" and ^ draw lines perpen- 
dicular to fg. From ^, along the line gk, set off the distances 
g, w, /, k, equal to the distances c, i, p, o (Fig. 258 a). From the 
points, m, /, and k^ draw lines parallel to fg till they meet the 
\\ne.fk. This completes the plan. 



140 



GEOMETRICAL DRAWING AND DESIGN. 



The heights K, H, G, which give the horizontal Hnes in the 
elevation, are obtained from the distances 6, z, d in the plan 
(Fig. 258 a). Having obtained these heights, draw the lines h'k' 
and n'm'. Carry up perpendicular lines from the points in the 
plan till they meet these lines, which give the corresponding 
points in the elevation. Join them as shown. 



187. 



To project a regular hexagonal prism 12 V' long, with sides 4" 
wide, resting on one of its smaller edges on the h.p. ; its longer 
edges to be inclined at an angle of 60° to the H.P., and parallel 
to the v.p. Its nearest edge to be 2" from v. p. Scale ^ full 
size. Fig. 259 A. 

In any convenient position draw the hexagon AEDF with 
4" sides, with lines joining the opposite angles as shown. Draw 

XY, and at any 
point d draw the 
line dc\ 12^" long 
at an angle of 60 
with XY. From a! 
and c' draw the 
lines ciU and c'd' 
perpendicular to 
a!c' . Set off the dis- 
tances a!^ /', b' 
along db\ equal to 
thedistancesE,B,F 
of hexagon. From 
these points draw 
lines parallel to cic' 
till they meet the 
line c'd' . This com- 
pletes the elevation. 
Draw a line 2" below XY, and parallel to it. From L on 
this line let fall a perpendicular, and on it set off the distances 
L, K, H, G equal to the distances A, B, C, D of hexagon. From 
each of these points draw lines parallel to XY, and let fall 
lines from the various points in the elevation till they meet 
these lines. This gives the corresponding points in the plan. 
Join them as shown. 




259- 



SIMPLE SOLIDS. 



141 



188. To project the same prism standing on one of its shorter 
edg-es on the H.P., at an angle of 30^ with v. p., with its longer 
edges inclined at an angle of 60° with the H.P., and in vertical 
planes inclined at an angle of 60° to v.p. ; its nearest comer to 
be 3" from v.p. Scale ^V full size. Fig. 259 B. 

Fix the position of point vi 3" below XY, and draw the line 
vip at an angle of 30° with XY. This represents the line KH in 
Fig. 259 A. The plan in Fig. 259 B is precisely the same as that 
shown in Fig. 259 A, turned to a different angle. Complete the 
plan against the line ;;//, from Fig. 259 A. As every point in the 
elevation is found in the same way, it is only necessary to 
describe one point. Erect a line on point r at right angles 
to XY till it meets a horizontal line drawn from point e' . This 
gives the point r. Find the other points in the same way, and 
join them, as shown. 

189. To project a regular hexagonal prism, Tj" long, with faces 2V 
wide, resting on one comer on the H.P. ; its longer edges to be 
inclined at an angle of 45° with the H.P.; one face to be parallel 
to the v.p. and 2V' from it. Scale \ full size. Fig. 260 .A.. 

Construct a hexagon with 2o" sides, and join the opposite 

J? 
, d.' 

k 




angles by lines at right angles to each other, as shown. Draw 
the line XY, and at any point (^ draw the line a!h\ 7\" long, at 
an angle of 45° with XY. From the points a' and h' draw the 



142 GEOMETRICAL DRAWING AND DESIGN. 

lines rtV and Hd' perpendicular to cih' . Set off the distances 
6', e\ f\ d on h'd' equal to A, B, C, D of hexagon, and draw lines 
from these points parallel to h'a' till they meet the line c^c. 
This completes the elevation. 

Draw the line mk 2V below XY. From H on mk produced 
draw the line HL perpendicular to mk, and set off the distances 
H, D, L equal to the distances E, B, F of hexagon. Draw Hnes 
from these points parallel to XY. All the points of the plan 
must come on these three lines, and are determined by dropping 
perpendiculars from the corresponding points in the elevation. 

190. To project the same solid, with its longer edges still inclined at 
an angle of 45° to the H.P. ; but instead of being parallel to the 
V.P., as in the last problem, let them be in vertical planes, 
inclined at an angle of 30° with the v. P. ; its nearest corner to 
the v.p. to be Ij" from it. Scale ^ full size. Fig. 260 B, 

Fix the position of point n ij" below XY, and draw the line 
no at an angle of 30° with it. The plan in this problem is 
precisely similar to the plan in the last problem, but turned round 
at an angle of 30° with XY, and the line no corresponds with 
the line km (Fig. 260 a). Complete the plan as shown. 

Every point in the elevation is found in the same manner. 
For example, erect a perpendicular on point o till it meets a 
horizontal line drawn from point e'; this will give the point o'. 
Proceed in the same way with all the other points, and join them. 

The Regrular Solids. 

There are five regular solids, and they are named after the 
number of faces they each possess ; viz. the Tetrahedron has four 
faces, the Hexahedron six faces, the Octahedron eight faces, the 
Dodecahedron twelve faces, and the Icosahedron twenty faces. 

They possess the following properties, viz. : 

(i) The faces of each solid are equal, and similar in shape, 
and its edges are of equal length. (2) All their faces are regular 
polygons. (3) All the angles " formed by the contiguous faces 
of each solid are equal. (4) Each can be inscribed in a sphere, 
so that all its angular points lie on the surface of the sphere. 

The student is advised to make these drawings to a scale 
three times that of the diagrams. 



SIMPLE SOLIDS. 



H3 



191. To project a tetrahedron with edges 9^' long-, with one face 
resting on the H.P. ; one of its edges to be at an angle of 16° 
with V.P., and its nearest angular point 3|" from v.P. Scale 
jV full size. Fig. 261 A. 

All the faces of this solid are equilateral triangles. 

Draw XY, and fix the position of point a 3I" below it. From 
a draw the line ah 
9" long, at an angle 
of 16° with XY. On 
the line ab con- 
struct an equilateral 
triangle ahc. Bisect 
each of the angles 
at ^, 6, and c by 
lines meeting at d. 
This completes the 
plan. 

To find the alti- 
tude of the eleva- 
tion, produce the 
line bd to e, and at 
d draw the line df 
perpendicular to eh. 
With b as centre, and radius be, draw an arc cutting d/mf. 

To draw the elevation, erect a perpendicular d'g' directly 
above d, and make d'g' equal to df. Carry up the points a, c, b till 
they meet XY in a', c', and b'. Join the point d' to a'^ c , and b'. 




Fig. 261. 



192. To project the same solid tilted on to one of its angular points 
with its base inclined at an angle of 20^ with H.P. Scale ^V full 
size. Fig. 261 B. 

Fix the point k' on XY, and draw the line k'k' at an angle 
of 20° with it, and equal in length to the line b'a' (Fig. 261 a). As 
this elevation is precisely similar to the last, complete it from 
that figure on the line k'k'. 

Draw lines from the points a, b, d, and c (Fig. 261 a) parallel 
to XY. All the points of the plan must come on these four 
lines, and are found by dropping lines at right angles to XY 
from the corresponding points in the elevation. 



144 



■GEOMETRICAL DRAWING AND DESIGN. 



193. To project a hexahedron or cube with edges 3|" long, resting 
on one edge on the H.P., and its base making an angle of 22° 
with it; the side nearest the V.P. to be parallel to, and l|" 
from it. Scale ^ full size. Fig. 262 A. 

All the faces of this solid are squares. 
Draw the line XY, and at any point d draw the line cih' at an 
angle of 22°. Complete the square dh'c'd' . 




Fig. 26 



Below XY draw the line ef i|" from it. Draw the lines a'a^ b'b, 
c'c and d'd at right angles to XY. Make eb and fd each equal 
to dh\ and join hd \ cc' cuts efm.g. 



194. To project the same solid with its edge stiU resting on the H.P. 
but inclined to the v. P. at an angle of 60°, the base still forming 
the same angle with the H.P., Tdz. 2°; its nearest corner to be 
1' from V.P. Scale \ full size. Fig. 262 B. 

Fix the position of point h \" below XY, and draw the line hk 
equal X.o fe (Fig. 262 a) at an angle of 30° with XY. Complete 
this plan from Fig. 262 A, to which it is precisely similar in shape. 

Draw lines through the points 5', </, and d (Fig. 262 a) parallel 
to XY. Draw perpendiculars from the points in the plan till 
they meet these lines, the intersections give the corresponding 
points in the elevation. Join these points as shown. 



SIMPLE SOLIDS. 



HS 



195. To project an octahedron, with edges 4" long, poised on one of 
its angiUar points on the H.P.; its axis to be perpendicular to 
H.P., and its edge nearest to the v.P. to be parallel to, and l|" 
from it. Scale ^ full size. Fig. 263 A. 

All the faces of this solid are equilateral triangles. 

Draw the line XY, and if" below it draw the line ab 4" long. 
Complete the square ^- 

ab(/c, and draw its di- 
agonals. This will be 
the plan of the solid. 

As every angular 
point is equidistant 
from the centre of a 
sphere circumscribing 
the solid, the point e 
must be the centre of 
the plan of the sphere 
and (u/ its diameter. 

Draw the projector 
ee' from the point t\ 
Make e'f equal in 
length to ac/. This will 
be the axis of the solid. 
Bisect e'/' by the line cUf parallel to XY, and erect perpendi- 
culars from the points c and ^/till they meet this line in the points 
c' and d. Join c'e' and c'f, d'e and d'f\ to complete the elevation. 

196. To project the same solid with one face resting on the H.P. ; 
the edge nearest the v.P. to be parallel to, and l|" from it. 
Scale \ full size. Fig. 263 B. 

Fix the positions of the points g' and // on XY the same 
distance apart as the points c' and/' in the preceding problem. 
On this line complete the elevation from Fig. 263 A. 

Draw lines from the points 6, e, and d parallel to XY. 
All the points of the plan must come on these three lines, 
and are found by dropping perpendiculars from their correspond- 
ing points in the elevation. 

The Dodecahedron and Icosahedron can hardly be described 
as ' simple solids.' Their projection will be found in Spanton's 
Complete Geometrical Course {Macinilla?i\ pp. 229 sqq. 

K 




146 



GEOMETRICAL DRAWING AND DESIGN. 



Octagonal Pyramids. 

197. To project a re^lar octagonal pyramid, 8" high, with each 
side of its base 2h" wide, standing on its base on the H.P., with an 
edge of its base parallel to the V.P. and 2^" from it. Scale jV full 
size. Fig. 264 A. 

Draw XY, and 2h" below it draw ad 2\" long. On ab 
construct a regular octagon, and join the opposite angles 




Carry up projectors from the points d, e^f^g perpendicular to 
XY. Fix the point c' immediately above r, and 8" above 
XY. Join the point c' to d'^ e\f\ and g\ which completes the 
pyramid. 



198. To project the same solid lying with one face on the H.P., but 
with its axis in a plane parallel to the v. P. Scale ^V full size. 
Fig. 264 B. 

On XY mark off the distance J^k' equal to g'c' (Fig. 264 a). 
Complete the construction of elevation from Fig. 264 A. 

Let fall lines from the various points of the elevation, at right 
angles to XY, till they meet lines drawn from the corresponding 
points in the plan (Fig. 264 a), parallel to XY. The intersection of 
these lines give the required points, by joining which we obtain 
the plan. 



SIMPLE SOLIDS. 



147 



199. To project the same solid resting on one of its shorter edges, 
■with its base inclined at an angle of 30^ with H.P. ; its axis to be 
in a plane parallel to the v. P. Scale ^ full size. Fig. 264 c. 

Draw XY, and fix the position of point o'. Draw o'p' at an 
angle of 30° with XY, and equal in length to /I'n' (Fig. 264 b). 
Complete the elevation from Fig. 264 b. 

In any convenient position draw the line DG perpendicular to 
XY, and set off upon it the distances D, E, F, G equal to /i\ /', 
m', n' (Fig. 264 b). Draw lines from these points parallel to XY. 
From the various points in the elevation let fall lines at right 
angles to XY till they meet these'lines, which give the correspond- 
ing points in the plan. 

Cones. 

Cones are projected in precisely the same way as polygonal 
pyramids. After finding the points in the base, instead of 
joining them by lines, as in the case of pyramids, a fair curve is 
drawn through them. Eight points of the base are found in the 
examples here given. Should more points be required, it is 
only necessary to select a pyramid having more sides than 
eight to construct the cone upon, 

200. To project a cone 8" high, with base 6h' in diameter, resting 
on the H.P. Scale yV full size. Fig. 265 A. 
Draw XY, and in any convenient position below it draw a 
circle 6V' in diameter, and a diameter ^/i parallel to XY. 

1' .V' 




Carry up projectors from g- and /i till they meet XY ; also 
from k, and produce ke 8" above XY. Join k'g' and ^7i\ 



148 GEOMETRICAL DRAWING AND DESIGN. 

201. To project the same solid resting on its edge, with its base 
inclined at an angle of 30^ with H.P. ; its axis to be in a vertical 
plane parallel to the V.P. Scale ^V fuU size. Fig. 265 B. 

Inscribe the circle (Fig. 265 a) in a square abcd^ the side 
nearest XY to be parallel to it. Draw diagonals, and through 
the centre draw diameters parallel to the sides of the square. 
Through the four points where the diagonals cut the circle 
draw lines parallel to the sides of the square. 

Fix the point m' on XY, and draw the line m'r' at an angle 
of 30° with it. Set off the- distances m'o'p'q'r' equal to the dis- 
tances /?', ;/, e\ l\g' (Fig. 265 a). Complete the elevation from 
Fig. 265 A. 

Let fall lines from the various points of the elevation at 
right angles to XY till they meet lines drawn from the corre- 
sponding points of the plan (Fig. 265 a). Draw a fair curve 
through the points forming the base, and lines from the vertex s 
tangential to the base. This completes the plan. 

202. To project the same solid, resting on its edge, with its base 
still inclined at an angle of 30° with H.P., but with its axis in a 
vertical plane, inclined at an angle of 60° with V.P, Scale j^Tr full 
size. Fig. 265 C. 

Draw the lines enclosing the base with the parallel lines 
intersecting each other where the diagonals cut the circle from 
the plan (Fig. 265 b). The line xy that passes through the axis 
to be inclined at an angle of 60° with XY. Complete the plan 
from Fig. 265 B. 

Draw lines from the various points of the plan at right 
angles to XY till they meet lines drawn from the corresponding 
points of the elevation (Fig. 265 b). Complete the elevation as 
shown. 



Cylinders. 

In the examples here given, only eight points of the circular 
base are projected, to save confusion of lines ; but any number 
of points can be found in the same manner. 



SIMPLE SOLIDS. 



149 



203. To project a cylinder, 5^" in diameter and 8j" high, standing- 
on its base on the h.p. Scale ^ full size. Fig. 266 A. 

Draw XY, and in any convenient position below it draw a 
circle 5V' in diameter. Draw tangents to it at the points aandd 
perpendicular to XY, 
and produce them 8j" 
above XY. Join the 
tops of these lines to 
complete the cylinder. 



204. To project the 
same cylinder, ly- 
ing on its side on 
the H.P., with its 
axis inclined at an 
angle of 45° with 
the V.P. Scale ^ 
full size. Fig. 266B. 

Draw four diameters 
tothe plan (Fig. 266 a), 
by first drawing the 
line ad parallel to XY, 
and then the other three diameters equidistant from it. This 
gives eight points in the circumference. Draw lines from these 
points parallel to XY, till they meet the line AE in the points 
A, B, C, D, E. 

At any convenient point c below XY (Fig. 266 b), draw the 
line cf at an angle of 45° with it. Draw the line cm, 8|" long, 
perpendicular to c/, and from m draw m£- parallel to cf. From 
m, along 7n^, set off the distances w, /, >^, //,^ equal to the distances 
A, B, C, D, E (Fig. 266 a), and from these points draw hues parallel 
to c?n. This completes the plan. 

Draw a line from / perpendicular to XY, and produce it 
above the ground line. Set off the distances E', D', C, B', A' on 
this line, equal to the distances E, D, C, B, A (Fig. 266 a), and 
draw lines from these points parallel to XY. Draw the pro- 
jectors from the various points in the plan till they meet these 
lines, which give projections of the points required to complete 
the elevation. 




Fig. 266. 



ISO 



GEOMETRICAL DRAWING AND DESIGN. 



205. To project the same solid, resting on its edge, with its base 
inclined at an ang-le of 30° with the H.P., its axis being parallel 
to the v.p. Scale ^ full size. Fig. 267 A. 

Draw XY, and at any point n' upon it draw the line n'r' 
at an angle of 30° with the ground line. On the line n'r' set 
off the distances n', o', p\ q\ r' equal to the distances g^ h, k. /, m 
(Fig. 266 b), and from each of these points draw perpendiculars 
to n'r\ 8j" long. Join s'^u'. This completes the elevation. 




From y let fall a line at right angles to XY, and set off upon 
it from any convenient point A the points B,C,D, E equal to the 
distances n\ o\ p\ q\ r'. From each of these points draw lines 
parallel to XY. From the various points in the elevation drop 
projectors till they meet these Hues in the corresponding points, 
by connecting which we get the plan. 

206. To project the same solid resting on its edge, with its base still 
inclined at an angle of 30° with the H.P., but with its axis inclined 
at an angle of 60° with the v.p. Scale ^ full size. Fig. 267 B. 

Draw the line FG equal to AE, inclined at an angle of 30° 
with XY. Complete the plan from Fig. 267 A. Draw projectors 
from the plan till they meet lines drawn parallel to XY from 
the corresponding points in the elevation (Fig. 267 a). These 
give the necessary projections for completing the elevation. 



SIMPLE SOLIDS. 



151 



Spheres. 

The plan and elevation of a sphere are circles ; but if we 
divide the sphere into divisions by lines upon its surface, such 
as meridians of longitude and parallels of latitude, we shall be 
enabled to fix its position and inclination to the co-ordinate 
planes, and project it accordingly. 

So as not to confuse the figure too much, we will restrict 
ourselves to eight meridians, with the equator, and two parallels 
of latitude. The junction of the meridians will of course give 
us the position of the poles, which will determine the axis. 

207. To project a sphere 5^" in diameter, with meridians and 
parallels ; its axis to be perpendicular to the H.P. 
size. Fig. 268 A. 

Draw XY, and 
in any conven- 
ient position be- 
low it draw a 
circle 5^" in dia- 
meter. Draw the 
diameters ab 
parallel to the 
ground line, and 
rtf/at right angles 
to it, and two 
other diameters 
equidistant from 
them. 

Produce the 
line dt above 
XY, and make 
c'p' equal in 

length to the diameter ab. Bisect c'p' in d' . With d as centre, 
and radius equal X.opa, draw a circle. Draw db' through d till it 
meets the circle in d and b' . Through d' draw the line e'f at 
an angle of 45° with db'., till it meets the circle in e and /'. 
From e' and /' draw lines parallel to db' till they meet the 
circle in g' and //. These lines represent parallels of latitude. 
Drop a perpendicular from g' till it meets ab in g. With p as 




Fig. 268. 



152 



GEOMETRICAL DRAWING AND DESIGN. 



centre, and radius pg^ draw a circle, 
parallel c'g' . 

v' 



This is the plan of the 




To avoid con- 
fusion, the pro 
jectors for half 
of one meridian 
only are shown ; 
but they are all 
found in the 
same manner. 

Erect a per- 
pendicular on 
point k till it 
meets the equa- 
tor in point k' ; 
also from point 
/till it meets the 
parallels in 
points /" and /'. 
Draw a curve 
which gives the projection of 



through the points c\ /", k\ /', p\ 

the meridian. 

208. To project the same sphere, with its axis inclined to the H.P. 

at an angle of 60°, but parallel to the V.P. Scale \ full size. 

Fig. 268 B. 

Note. — The same letters are taken throughout these spherical 
problems to facilitate reference. 

Draw a circle 5^" in diameter, touching XY, and draw the 
line c'p' at an angle of 60" with it. Draw the line db' at right 
angles to c'p\ and set off the distances of the parallels above and 
below a!b' equal to their distances in the elevation (Fig. 268 A). 
Draw the lines e'g and hf parallel to alb'. Complete the 
elevation from Fig. 268 A. Draw lines from all the points of 
intersection between the meridians and parallels of the plan 
(Fig. 268 a) parallel to XY, and let fall perpendiculars from the 
corresponding points in the elevation till they meet these lines, 
which give the projections of the points of intersection. Draw 
the curves. 



SIMPLE SOLIDS. 



153 



209. To project the same sphere, with its axis still inclined to the 
H.P., at an angle of 60°, hut in a 
vertical plane inclined at an angle of 
60° with the V.P. Scale ^ full size. 

Fig. 269. 

Draw the line cp at an angle of 60° 
with XY, for the plan of the axis, and on 
this line complete the plan from Fig. 268 B. 
Perpendicular to XY draw the line XL, 
and set off the distances X, C, L, K, C, L 
equal to the distances Y, C, L, K, C, L 
(Fig. 268 b). From each of these points 
draw lines parallel to XY till they meet 
projectors drawn from the corresponding 
points in the plan, which give the projec- 
tions required. 

Fig. 269. 




EXERCISES. 

1. The plan is shown of three bricks (Fig. 270), each 9" x 4^" x 3", 
one resting upon the other two. Draw an elevation upon the given 
line xy. 

Scale (which need not be drawn) 2" to i', or \ of full size. (April, '98.) 




Fig. 270. 




Fig. 271 



2. The plan is given (Fig. 271) of a flight of three steps each |" 
high, of which 3 is uppermost. Draw an elevation on the given xy. 

(June, '00.) 



J,54 



GEOMETRICAL DRAWING AND DESIGN. 



3. Plan and elevation are given of a solid letter H (Fig. 272). Draw 
an elevation when the horizontal edge ab makes an angle of 60 with the 
vertical plane of projection. (June, '97.) 



G, b 




A 

Fig. 272. Fig. 273. 

4. The diagram ( Fig. 273) shows the plan of two square prisms, one 
resting upon the other. Draw their elevation upon the given xy. The 
lines AB and CD are plans of square surfaces. (June, '99.) 

5. The plan is given of a cube (Fig, 274), having a cylindrical hole 
pierced through its centre. A vertical plane, represented by the line Im, 






+ 

IlevcUion 



Fig. 274. Fig. 275. 

cuts off a portion of the solid. Draw an elevation on the line xy, supposing 
the part of the solid in front of Im to be re- 
moved. The part in section should be clearly 
indicated by lightly shading it. (April, '96.) 

6. The plan is given (Fig. 275) of a right 
prism having equilateral triangles for its bases. 
These bases are vertical. Draw an elevation 
of the prism on the line xy. Show the form 
of the section of the prism made by the vertical 
plane Im. The part in section should be indi- 
cated by lightly shading it. (June, '98.) 

7. Plan and elevation are given (Fig. 276) 
of a solid composed of a half-cylinder placed 

f'ig- 276. upon a prism. Draw a new elevation, when 

the horizontal edges of the prism make angles of 45° with the vertical 
plane of projection. (April, '96. ) 




SIMPLE SOLIDS. 



^55 



8. An elevation is given (Fig. 277) of an archway with semi-circular 
head, in a wall i' 6" thick. Draw a second elevation upon a vertical 
plane which makes an angle of 45° with the face of the wall. (Scale, 
which need not be drawn, ^' to i'.) (June, '00.) 





Fig. 277. 



Fig. 278. 



Fig. 279 



9. Plan and elevation are given (Fig. 278) of a rectangular block with a 
semi-cylindrical hollow in it. Draw a new elevation upon a vertical plane 
which makes an angle of 30° with the horizontal edge ab. (April, '98.) 
10. The diagram (Fig. 279) shows a side elevation of a square prism, 
pierced through its centre by a cylinder. Draw a front elevation of 
the solids. (June, 'cxd. ) 



Elevation 




PlOJl 

Fig. 280. 



Fig. 281. 



11. Plan and elevation are given (Fig. 280) of a cylinder through 
which a square opening has been cut. Draw a fresh plan and elevation 
of the solid, the plane of the circular base FG being inclined at 45° to 
the vertical plane of projection. (June, '99.) 

12. The "block" plan and end elevation are given (Fig. 281) of a 
building having a square tower with pyramidal roof. A side elevation 
of the building is required. (June, '98.) 



156 



GEOMETRICAL DRAWING AND DESIGN. 



13. The plan is given of a piece of cylindrical rod (Fig. 282) cut by a 
vertical plane shown at Im. Draw an elevation of the solid upon an xy 
parallel to Im. (April, '99.) 





Fig. 282. 



Fig. 283, 



14. Plan and elevation are given of a sloping desk (Fig. 283). Draw 
an elevation upon a vertical plane parallel to the Xvtxo. pq. Show upon 
this elevation the outline of the section made by the vertical plane 
represented by /</. (April, '99.) 

15. The diagram (Fig. 284) shows the plan of two cubes, one resting 
upon the two others, with a sphere resting on the upper cube. Draw 
an elevation on the given xy. (April, '00.) 





Fig. 284. 



Fig. 285. 



16. Show in plan and elevation a shallow circular metal bath. 
Diameter at top 2' 6", at bottom 2', height 6". The thickness of the 
metal may be neglected. Scale (which need not be drawn) i' to \". 

(April, '99.) 

17. The diagram (Fig. 285) shows the elevation of a right cone having 
its vertex at V. Draw the plan. (April, '00.) 



CHAPTER XIV. 

ORTHOGRAPHIC PROJECTION. 



Preparatory to the study of sections of solids it is desirable 
to have a more thorough insight into the principles of Ortho- 
graphic Projection, though its simpler applications need only be 
considered. 

In Solid Geometry objects are projected by means of parallel 
projectors perpendicular to 
two co-ordinate planes. These 
planes may be considered as in- 
definite in extent. For instance, 
the H.P. might be extended 
beyond the V.P., and the v. p. 
below the H.P. 

To understand this fully, let 
us take two pieces of cardboard 
about 12" square, and half-way 
across the middle of each cut a 
groove, as shown in Fig. 286. 
By fitting these two pieces 
together we obtain two planes intersecting each other at right 
angles, as shown in Fig. 287. 

We have now four sets of co-ordinate planes, forming four 
" dihedral angles," identified by the letters A, B, C, D. 

The angle formed by the upper surface of the H.P. with the 
front of the V.P. is called the " first dihedral angle," viz. A, fig. 287. 

The angle formed by the upper surface of the H.P. with the 
back of the V.P. is called the " second dihedral angle," viz. B, fig. 287. 

The angle formed by the under surface of the H.P. with the 
back of the v.p. is called the "third dihedral angle," viz. C, fig. 287. 

The angle formed by the under surface of the H.P. with the 
front of the V.P. is called the " fourth dihedral angle," viz. D, fig. 287. 




158 



GEOMETRICAL DRAWING AND DESIGN. 




We will now take a piece of cardboard 4" x 3", and place one 
of its shorter edges against the H.P. and a longer edge against 

the v.p. in the first 
dihedral angle, with 
its surface perpen- 
dicular to each 
plane (Fig. 287). 
Let the corner A 
represent a point 
we wish to project 
on to each plane : 
the top edge Aa' 
represents its verti- 
cal projector, and 
the point a! its ver- 
tical projection; the 
edge Ka represents 
its horizontal projector, and the point a its horizontal projection, 
and so on, placing the cardboard in each of the dihedral angles. 

Lines. 

To illustrate the projection of lines, we will restrict ourselves to 

the two co-ordinate planes 
of the first dihedral angle 
(Fig. 288). 

Take the piece of card- 
board and place it with 
one of its shorter edges 
on the H.P., with its sur- 
face parallel to the v.p. 
Let the top edge AB 
represent a line we wish 
to project. The edges ha 
and B6 will then represent the horizontal projectors, and the line 
ah its horizontal projection. If we draw lines A«' and B6' 
perpendicular to the V.P. from the points A and B, they represent 
the vertical projectors, and the line clU its vertical projection. 

We will now place the piece of cardboard touching both 
planes, with one of its shorter edges on the H.P., and its surface 




ORTHOGRAPHIC PROJECTION, 



r?9 



perpendicular to both planes. Let the edge Cc represent the 
line to be projected, Cc" and cc' represent the vertical projectors, 
and the line c'c" its vertical projection. The point c on the H.P. 
is called the " horizontal trace " of the line. 

Note. — The point where a line, or a line produced, would 
meet either plane is called the " trace " of that line : if this point 
is on the H.P., it is called the "horizontal trace" (h.t.); and if it 
is on the v.p., the "vertical trace" (v.T.). The same thing 
applies to the projection of planes. 

Now place the piece of cardboard with one of its longer edges 
on the H.P., and its surface per- 
pendicular to both planes. Let 
the top edge Dd" represent the 
line to be projected. The edges 
D^and ^W represent the hori- 
zontal projectors, and the line X 
dd' its horizontal projection. 
The point d" is its vertical trace. 

Fig. 289 represents the co- 
ordinate planes opened out 
into one flat surface. The pro- 
jections below XY represent Fig. 289. 
the plans of the lines, and those above XY the elevations. 

We will now use the same piece of cardboard to illustrate the 
projections of lines inclined to one or both co-ordinate planes 



a 


b 


,c ' 

i / 

■ 








\c' 


d' 




b 


i 




a 






^0 


d 




Fig. 290. 

(Fig. 290). In the first case we will incline it to both planes, with 
one of its shorter edges resting on the H.P. and parallel to the v,P 



i6o 



GEOMETRICAL DRAWING AND DESIGN. 



Let the edge AB represent the Hne to be projected. aB is its 
horizontal, and a'd' its vertical projections. 

Then incline it to the H.P., with one of its shorter edges still 
on the H.P., but perpendicular to the v.P. 

Let CD represent the line to be projected. The line D^: is 
the horizontal, and c'ci' its vertical projections. 

Now incline it to the V.P., with its lower longer edge parallel 
to, but raised a little above the H.P, 

Let EF represent the line to be projected. The line e/ is its 
horizontal, and/V its vertical projections. 

Let us now draw a diagonal GF across the piece of cardboard 
and again hold it in the same position ; and let GF represent 
the line to be projected. 

The line e/ still represents its horizontal projection, but the 
line/'^' is its projection on the vertical plane. 

Fig. 291 shows the plans and elevations of these Hnes, with the 
co-ordinate planes opened out flat. 

We have now projected a line in seven distinct positions, 
viz. : 
Fig. 288. — AB parallel to H.P. and parallel to 

C perpendicular to „ „ 

D parallel to , 



Fig. 290. 



-AB inclined to 
CD 

EF parallel to 
GF inchned to 





a' 


/ 


c ' 

/ 


/ 1 ' 




b' 


V 








a 






/ 




R 


i 




/ 




Pi 


' f 


/ 



Fig. 291. 



perpendicular to 

inclined to 
parallel to 
inclined to 



The student should parti- 
cularly notice the difference 
between AB and FG in 
Fig. 290. Although they 
are both inclined to both 
planes, AB is in a vertical 
plane perpendicular to the 
V.P., while FG is in one 
inclined to the v.p. 

We will now project these 
lines in the various positions 
to scale. 



ORTHOGRAPHIC PROTECTION. 



i6i 



210. To project a line AB 2^" long, parallel to botli the H.P. and 
V.P., its distances to be 3" from the H.P. and ih" from, the V.P. 
Scale J full size. Fig. 292. 

Draw XY, and iV' below it draw the line a^ 2^' long. Draw 
the projectors aa' and bb' at right angles to XY, and 3" above it. 
Join a'b'. 

211. To project a line CD 3|" long to the same Scale, parallel to 
the V.P. and 2^" from it, but perpendicular to the H.P. Fig. 292. 

Fix the position of point c 2^" below XY. Draw a line per- 
pendicular to XY, and produce the same 3I" above it. c is the 
plan or H. trace, and c'd' the elevation required. 




212. To project a line EF 3" long to the same scale, parallel to the 
H.P. and 2j" above it, but perpendicular to the v. p. Fig. 292. 

Below XY, and perpendicular to it, draw the line ef 3" long. 
Draw the projector /V 2^" long, ef is the plan, and e' the 
elevation or V. trace. 

213. To project a line GH 3" long to the same scale, parallel to the 
V.P. and 1^" from it, but inclined to the H.P. at an angle of 60°o 
Fig. 292. " 

At any point £-' on XY draw the line g^/t' 3" long, and inclined 
to the H.P. at an angle of 60°. Let fall the projectors o^'o and 
A'A at right angles to XY. Set off the points ^ and /i i^" below 
XY, and join them. 

214. To project a line KL 3 " long to the same scale, inclined to the 
H.P. at an angle of 60°, but in a vertical plane perpendicular to 

the V.P. Fig. 292. 
Draw the line k7 at right angles to XY till it meets horizontal 
Hnes drawn from // and //. k'l' is the elevation, and /'/the plan. 

L 



J62 



GEOMETRICAL DRAWING AND DESIGN. 



215. To project a line MN 3" long to the same scale, parallel to the 
H.P. and l|" atoove it, but inclined to tlie v. p. at an angle of 
45°. The end of the line nearer the V.P. to be |" from it. 
Fig. 292. 



Fix the point n |" below XY, and draw 



5" long at an 



angle of 45° with it. Carry up the projectors perpendicular to 
XY, and produce them i^" above it in the points in' and ;/. 
Join in'7i' . 

216. To project a line OP 3" long- to the same scale, inclined to the 
H.P. at an angle of 30°, but in a vertical plane inclined to the 
V.P. at an angle of 60 ; one end of the line to be on XY. 
Fig. 292. 

From point o' on XY draw a line o'K 3" long, and inclined 
to XY at an angle of 60°. From the same point o' draw the 
line ^'B at an angle of 30° with XY. With o' as centre, and 
radius <?'A, draw an arc till it meets ^'B in B. Draw the Hne 
BC perpendicular to XY. With 0' as centre, and radius o'C^ 
draw an arc till it meets o' k. mp. Draw the projector pp' till 
it meets a horizontal line drawn from B in p' . Join p'o'. op 
is the plan, and op' the elevation of the line required. 

Planes. 

The lines in which planes intersect the co-ordinate planes 
are called traces : if on the H.P., the horizontal trace (h.t.) : 




Fig. 293. 

and on the v.p., the vertical trace (v.t.). The inclination of 
planes is determined by means of these traces. 



ORTHOGRAPHIC PROJECTION. 



163 



We will take the same piece of cardboard that we have used 
for our previous illustrations and place it on the H.P. and 
parallel to the V.P., as A (Fig. 293), The line ab^ where it 
intersects the H.P., will be its H.T. 

If we place it parallel to the H.P. and perpendicular to the v. p., 
as B, the line </</, where 
it intersects the v. p., 
will be its v.t. 

By placing it perpen- 
dicular to each plane, as 
C, ^will be its H.T. and X 
eg its V.T. 

On opening these co- 
ordinate planes out flat 
these traces will appear 
as shown in Fig. 294. 

We will now place the 
piece of cardboard perpendicular to the H.P. and inclined to 
the V.P., as D (Fig. 295) : hk will then be the H.T., and ///' 
the V.T. 





Fig. 295. 



Now incline it to the H.P. and make it perpendicular to the 
V.P., as E (Fig. 295) : mn will be its H.T. and mo' its V.T. 

By inclining it to both planes with its shorter edges parallel 
to XY, as F (Fig. 295), pq will be the H.T. and r's' the V.T. 



164 



GEOMETRICAL DRAWING AND DESIGN. 



For our next illustration we will take a 60° set- square G, as 
a right angle will not fit closely to the two planes in this posi- 
tion ; tu will be the H.T. and tw' the V.T. 

If we now open the planes as before, these traces will be 
shown as in Fig. 296. 





, 






0' 


r' s' 


/o^ 


w' 


/ 


t^ 






/ 


/ 








aW 


m 










t^ 


^ 


/ 

5 


1 


■) q 




\ 













Fig. 296. 



From these illustrations we can 'deduce the following facts — 

A plane can have no trace on the plane it is parallel to (see 
A and B, Fig. 293. 

If traces are not parallel to XY, they must intersect each 
other on that line (see D, E, and G, Fig. 295). 

If the traces of a plane are in one straight line when the H.P. 
and v.P. are opened out so as to form one continuous surface, 
the angles the plane forms with each co-ordinate plane must 
be equal. 

When a plane is perpendicular to either co-ordinate plane, 
its inclined trace will always give the amount of its inclination 
to the other co-ordinate plane ; e.g. hk (Fig. 296) forms with XY 
the angle ^ or the inclination of D (Fig. 295) to the v.P., while 
mo' forms with XY the angle d or the inclination of E to the H.P. 

When a plane is inclined to both planes, but has its traces 
parallel to XY, the sum of its inclinations, i.e. d+cfi = go°; as 
F (Fig. 295). 

The traces of planes inclined to one or both planes are not 
supposed to finish at XY ; they are indefinite, and are generally 
produced a little beyond XY. 



ORTHOGRAPHIC PROJECTION. 



165 



217. To find the traces of the foUowing planes. Scale ^ full size. 

Fig. 297. 

A,3" X 2^" perpendicular to the H.P. and inclined to the v.P. at 60°. 
B,3rxif „ „ v.P. „ „ H.P. „ 45°- 

C,4^"X3" inclined „ H.P. at 60° with shorter edges 

parallel to each plane. 

Draw XY ; and at any convenient point a draw a6 3" long, 
and at an angle of 60° with XY. From a draw ac' 2^" long. 
Then ad is the H.T., and ac' the v.T. of A. 




Fig. 297. 



From any convenient point <-/ draw ^/' 3I" long, and at an 
angle of 45° with XY. From d draw de if" long perpendicular 
to xy. Then de is the H.T., and t/f the V.T. of B. 

Take any point ^ on XY, and draw o-/i' 4I" long at an angle 
of 60° with it. From // draw /I'k perpendicular to XY ; and 
from o- as centre, with radius ^k, draw an arc till it meets a 
perpendicular from o- in /. From g- draw a perpendicular till it 
meets a horizontal line from // in 7/1'. Draw ;;/'// and /o, each 
3" long, parallel to XY. Then 7/1' 72' will be the v.T., and 0/ the 
H.T. of C. 

We will now proceed with planes that are inclined to both 
planes of projection : they are called oblique planes. Let us take 



1 66 



GEOMETRICAL DRAWING AND DESIGN. 



a 60° set square and place it so as to fit closely against both 
planes, as shown at A (Fig. 298). ca will be the H.T. and cU 
the v.T. 

The inclination of a plane to the coordinate plane containing 
its trace is the angle between two lines perpendicular to the 
trace, one in the co-ordinate plane and one in the plane itself. 




Fig. 298. 

The line ca (Fig. 298) is the H.T. of the plane A, and ab' is 
a line in the plane A, and af a hne in the H.P., both perpen- 
dicular to the H.T.; therefore b'af is the angle A forms with 
the H.P. 



218. To determine the traces of a plane inclined at an angle of 45° 

to the H.P., and at an angle of 35° to the v.P. Fig. 299. 

Note. — In the Definitions (page 10) a cone is described as 
being generated by the revolution of a right-angled triangle 
about one of its sides as an axis. The hypotenuse of this 
triangle is called a generatrix. 

The problem is generally solved in the following manner : 
The generatrices of two cones forming the necessary angles to 
the two planes of projection are determined with their axes 
meeting at the same point on XY. The sides of these two 
cones should be tangential to a sphere, the centre of which is the 



ORTHOGRAPHIC PROJECTION. 167 

point on XY in which their axes meet. The plane required is 
tangential to the bases of these two cones. 

Draw XY. Select any point c for the point in which the 
axes of the cones meet, and draw a line through it at right 
angles to XY, At any point d on XY draw a line at an angle 
of 60° with it till it meets the perpendicular on c in b' . With c 
as centre, and radius cd^ draw the semicircle def. Join fb'. 
Then def\% the plan, and /^W the elevation of a semi-cone. 



From c draw the line eg perpendicular to dU . With c as 
centre, and radius cg'^ draw a circle. This will represent the 
plan and elevation of a quarter of the enveloped sphere. 

Draw the line ah^ at an angle of 45°, tangential to the 
plan of the sphere, cutting b'c produced in a. With c as 
centre, and radius ch^ draw the semicircle hk'l. Join al. 
Then hal will be the plan, and Ik'h the elevation of another 
semi-cone. 

From a draw the line m7i tangential to the semicircle y^^; 
i.e. the base of the horizontal semi-cone. 

From b' draw the line b'ln tangential to the semicircle Ik'h \ 
i.e. the base of the vertical semi-cone. 

Then a7n is the H.T. and b'm the V.T. required. 



i68 GEOMETRICAL DRAWING AND DESIGN. 

Fig. 300 is a perspective view showing this construction. The 
horizontal semi-cone is dotted in each instance. 

There is another method of finding the traces for an oblique 
plane, viz. by first finding the projections of a line perpendicular 




Fig. 300. 

to the plane required, and then drawing the traces at right 
angles to these projections. This will be more easily understood 
by referring to the set-square B (Fig. 298). 

Let Op represent a line at right angles to on\ and perpen- 
dicular to the plane B. Then op is the horizontal projection, 
and o'p the vertical projection of this line {Op) ; and the H.T. ;;?<?, 
and the V.T. inti ^ are at right angles to these two projections. 



{For Exercises see p. 226,) 



CHAPTER XV. 

SECTIONS OF SOLIDS, CONSTRUCTION OF SECTIONAL 
AREAS. 

A section is defined as the intersection of a solid by a plane. 
This plane is called the cutting plane, and in the following 
problems it is given inclined at different angles to both the 
co-ordinate planes. The surface of the solids cut through are 
projected, and the true shapes of the sectional area are 
" constructed." 



219. To project a cube of V edge, standing on the H.P., and inclined 
at an angle of 30 to the v. P., intersected by a cutting plane 
inclined to the H. P. at an 
angle of 45", and perpendi- 
cular to the V.P. ; the plane 
to intersect both the hori- 
zontal faces of the cube. 
Fig. 301. 

Draw the plan adcd of the 
cube, and carry up projectors 
from the points, |" above XY, 
and join them for the elevation. 

Find the traces of the cutting 
plane (Prob. 218). 

Where the v.T. cuts the ele- 
vation in the points e' and /', 
drop projectors which will in- 
tersect the plan in the lines e/i ' 
and /£. afgche is the plan of 
the cut surface of the cube, and e'a'c'f the elevation. 




I70 



GEOMETRICAL DRAWING AND DESIGN. 



The sectional surface can be " constructed ■' by rotating thf 
projecting surface of the section on either the h.t. or V.T. 

To rotate the plan of the section on the H.T. Draw lines at 
right angles to H.T. from the points of the plan, and make the 
lengths of these lines from the H.T. equal to the distances of 
the corresponding points on the V.T. from XY ; e.g. to obtain 
the point C, set off /tC equal to e'c\ and so on with each of the 
other points. Join them, as shown, to complete the construction l 
of the sectional surface. 

To rotate the sectional surface on the V.T. Draw the perpen- 
diculars from the points e\ a\ ^, /', and make the lengths of 
these lines equal to the distances of the corresponding points 
below XY ; e.g. make/'G' equal to mg.,f'Y' equal to inf., and so 
on with the other points. Join them as shown. 

220. To project a quadrilateral prism 10 V' long, with a base 6" 
square, standing on its base on the H.P., with its longer edges 
parallel to the v. p., and one of its sides inclined to the v. p. at 
an angle of 60^ Intersect the prism with a plane parallel to 

XY, and inclined to 
the H.P. at an angle 
of 30'. Scale J^ full 
size. Fig. 302. 

Project the prism 
(Prob. 171). Find the 
traces of the cutting 
plane (Prob. 218). 

At any convenient 
point e draw the line 
E/ at right angles to 
XY cutting the H. and 
V. traces. With e as 
centre, and radius ^E, 
draw an arc till it meets 
XY vcif. Join e'f. efe' 




Fig. 302. 



is the angle the cutting plane forms with the H.P. 

Draw lines from each point in the plan perpendicular to ^E, 
and meeting it in the points C, B, D, A. With e as centre, and 
each of these points as radii, draw arcs ; and where they meet XY 
draw perpendiculars till they meet ^yin the points C', B', D', A'. 



SECTIONS OF SOLIDS. 



171 



From these points draw horizontal hnes till they meet projectors 
drawn from the corresponding points in the plan in the points 
c\ b\ d^ a'. Join these points for the elevation of the section. 

To construct the sectional area. Draw perpendiculars to e'/aX 
the points C, B', D', A', and make their lengths equal to the 
distances of the corresponding points in the plan from the line 
^E ; e.g. make C'C" equal to Cf, B'B" equal to B<^, and so on 
with the other points, and join them. 

221. Project a regular hexag-onal prism 9V' long, with Stt" sides, 
standing on its base on the H.P., with one of its faces inclined 
to the V.P. at an angle of 58 . Intersect the prism by a plane 
inclined at an angle of 55' with the H.P., and 46 with the v.p., 
the plane to cut through the base of the prism. Construct on 
the H.P. the sectional area. Scale ^.i full size. Fig. 303. 

Project the prism (Prob. 185). P^ind traces of cutting plane. 

Draw lines from the points of the plan parallel to the H.T. 
till they meet XY, then draw perpendiculars to XY, till they 
meet the V.T. in the points 
C, A', D', F', E'. From 
these points draw horizontal 
lines till they meet projectors 
drawn from the correspond- 
ing points of the plan in the 
points c\ d\ e\f\ a\ b\ and 
join them as shown. This is 
the elevation of the section. 

To construct the true shape 
of the section, we rotate the 
plan abcdef on the H.T. From 
e draw the line eg parallel to 
the H.T., and equal in length 
to the height of ^'above XY. 

Draw lines from each of 
the points in the plan at right 
angles to the H.T. : the one 
drawn from e will intersect the H.T. in //. With this point as 
centre and the radius kg., draw an arc till it meets the line from 
e produced in E. Find all the other points in the same way as 
shown. 




Fig. 303. 



172 



GEOMETRICAL DRAWING AND DESIGN. 



The student should observe that the Hue JiK is the true 
length of the line he. A simpler method of obtaining the 
length of these lines is as follows. 

Note. — The true length of a projected Hne is equal to the 
hypotenuse of a right-angled triangle, the base of which is 
one of its projections, and the altitude the perpendicular height 
of the other projection. 

Let kd represent the horizontal projection of a line, and Yid' 
its perpendicular height. From k set off on the H.T. k7J equal 
to Kd'. Then the distance between 7i and d will be the true 
length of the line kd. Set this distance off from k on dk 
produced. This will give the point D. The other points 
can be found in the same manner. 



222. Project a regular hexagonal prism 10 V' long-, with 3^ 

lying on one of its faces on the H.P., with its longer edges 
inclined to the v.p. at an angle of 17°. Intersect the prism 
with a plane inclined at an angle of 50° to the h.p. and 56° to 

the v.p. ; the plane to 
intersect all its longer 
edges. Construct on the 
v.p. its sectional area. 
Scale jV full size. Fig. 
304- 
Project the prism (Prob. 
186). 

Find the traces of the 
cutting plane (Prob. 218). 

Let BC be the side of the 
prism resting on the H.P. 
Join AD. Then BC, AD, 
and EF are horizontal lines, 
and B'A' and B'F' their 
heights above XY. 

Note. — A horizontal Hne 
contained by a plane would 
Fig. 304- be drawn parallel to the H.T. 

on plan, and parallel to XY in elevation. 

As be is on the H.P., it must coincide with the H.T. Where 
the lines F' and A' produced meet the V.T., drop perpendiculars 




SECTIONS OF SOLIDS. 173 

till they meet XY, and then draw lines parallel to the H.T. 
Where these lines intersect the lines of the plan they will 
determine the points of the section. Join them as shown. 

Carry up projectors from these points till they meet the 
corresponding lines in the elevation, and join them. 

Note. — The projectors are omitted in several of these 
problems to save confusion, but the points in plan and eleva- 
tion bear corresponding letters throughout, so they can be easily 
recognised. 

To construct the sectional area on the V.P. Draw lines from 
each of the points in the elevation at right angles to the v.T. 
Take the distance of point e below XY as ^E', and set it off on 
the v.T. from g as gh. Set off gY!' equal to he\ the hypotenuse 
of a right-angled triangle, as described in the preceding problem. 
Find the other points in the same way, and join them as shown. 

223. Project a regular pjrramid standing on its base on the H.P., 
with its sides inclined to the V.P. at an angle of 45' ; the cutting- 
plane to be inclined at an angle of 43^ to the H.P. and 70° to 
the V.P. Construct the sectional area on the V.P. Fig. 305. 

Project the pyramid (Fig. 248). 

Find the traces of the cutting plane (Prob. 218). 

Produce the diagonal eg till it meets the H.T. in E. Draw 
the projector EE', and from E' draw a line parallel to the V.T. 
Where this line meets the edges of the pyramid in b' and d' will 
determine two points in the section. Drop projectors from 
these points till they meet the corresponding hues of the plan 
in the points b and d. 

The lines forming the section of this pyramid are really the 
intersection lines of two planes ; the cutting plane being one, 
and each side of the pyramid the other plane. We know that 
the line of intersection between two planes must have its .trace 
where the traces of the two planes intersect. Produce three 
sides of the base of the pyramid till they meet the H.T. in 
the points k^ m, and /. These are the traces of the lines 
required. Draw a line from k through b^ and produce it till 
it meets the diagonal fh in a. Draw lines from jn and / in 
the same manner till they meet the diagonals in d and c. 
Join dc. This will complete the plan of the section. 



174 



GEOMETRICAL DRAWING AND DESIGN. 



To determine the elevation of the points a and c. With 
centre o^ and radii oa and oc, draw arcs till they meet the 
diagonal es[ in the points 7t and/. Draw projectors to these 
points till they meet the edges of the pyramid in the points n' 




Fig. 305. 

and p'. Draw horizontal lines from these points till they meet 
the other edges of the pyramid in the points c' and a!. Join 
the points as shown, to complete the elevation of the section. 

Another method of obtaining the section of this pyramid is 
to assume a horizontal line q'r' in any convenient position in 
the elevation, and drop a projector from r' till it meets the 



SECTIONS OF SOLIDS. 



175 



diagonal eg in r. Draw rt parallel to the base gh. Produce 
q'r' till it meets the v.T. in /. Draw a projector from / till it 
meets XY in s. Draw a line from s parallel to the H.T. till it 
meets the line r/ in 11. Then ii is a point in the plan of the 
section, which can be completed from the traces k, ;;/, and /, as 
previously described. 

To construct the sectional area ABCD, proceed in the manner 
described in the preceding problem. 



224. To project a section through a right vertical cone : the cutting 
plane to be perpendicular to the v. P. ; to be inclined at an 
angle of 45° to the axis of the cone, but not to intersect its 
base. Fig. 306. This section is an ellipse. 

Let DE be the elevation of the section. Divide it into any 



Draw the axis of the cone, 

'% 
•if 



number of equal parts, e.g. six. 
and through the divisions on DE 
draw lines parallel to the base of 
the cone. 

The plan of the section is 
determined by first finding a 
succession of points in the curve, 
and then drawing a fair curve 
through them. We will take the 
points dd as an example. With 
C as centre, and a radius equal 
to G'H {/.e. the radius of the cone 
at the level of &'), draw an arc till 
it meets a projector drawn from 
d' in the points dd. Proceed in 
the same manner with the other 
points, and draw a fair curve 
through them. Fig. 306. 

To construct the sectional area. Draw the line D'E' in any 
convenient position, parallel to DE, and draw hnes from each 
of the divisions on DE at right angles to D'E'. Take the 
distance Gd from plan, and set it off on each side of G" in the 
points d"d". Find all the other points in the same manner, and 
draw a fair curve through them. 




176 



GEOMETRICAL DRAWING AND DESIGN. 



225. To project a section through a right vertical cone ; the cutting 
plane to be parallel to the side of the cone and perpendicular 
to the V.P. Fig. 307. This section is a parabola. 

Let D'E' be the elevation of the section. Divide it into any 

number of parts — it is better 
to have the divisions closer 
together towards the top. 
Draw horizontal lines through 
these divisions. 

The plan is determined by 
finding a succession of points 
as in the preceding problem. 
We will take the points bb 
as an example. 

With C as centre, and a 
radius equal to the semi- 
diameter of the cone at the 
level of the division b\ i.e. 
G'H, draw an arc till it meets 
a projector from b' in the 

points bb. Find the other points in the same manner, and draw 

a fair curve through them. 

To construct the sectional area. Draw the hne D"E" in any 

convenient position parallel to D'E', and draw lines at right 

angles to it from the divisions on D'E'. Take the distance 

Qb from plan, and set it off on each side of G" in the points b"b". 

Proceed in the same manner with all the other points, and draw 

a fair curve through them. 




Fig. 307. 



226. To project a section through a right vertical cone; the cutting 
plane to be perpendicular to the H.P, and inclined at an angle 
of 50° to the V.P. Fig. 308. This section is an hyperbola. 

Let DE be the plan of the section. From C draw the line 
(Zd perpendicular to DE. With C as centre, and radius Qd^ 
draw an arc cutting AB in n. Draw the projector nn' . n' is 
the vertex of the section. Divide the height g'ji' into any 
number of divisions, — they should be made closer together near 
the vertex, — and draw horizontal lines through them till they 
meet the sides of the cone. 



SECTIONS OF SOLIDS. 



77 



The elevation of the section is determined by first finding a 
succession of points, and then drawing a fair curve through 
them. We will take the points b'b' as an example. 

With the point C on plan as centre, and a radius equal to 
G'H (the semi-diameter of the cone at the level of b'\ draw arcs 




intersecting the line DE in the points bb. Draw projectors to 
these points till they meet the line drawn through G'H in the 
points b'b' . Find the other points in the same manner, and 
draw a fair curve through them. 

If the cutting plane were perpendicular to both the co-ordinate 
planes, gh would be the plan, and g'ji' the elevation of the 
section of the cone. 

To construct the sectional area. Draw the line ii'g" in any 
convenient position parallel to the axis of the cone. Produce 



178 



GEOMETRICAL DRAWING AND DESIGN. 



the divisions on 7ig' . Take the distance db from plan, and set 
it off on each side of G" in the points b"b". Find all the other 
points in the same way, and draw a fair curve through them. 

As the three preceding problems are conic sections, their 
sectional areas could be constructed by the methods described 
in Chap. XI. (Plane Geometry), but we must first determine 
the major and minor axes of the ellipse, and the directrices and 
foci of the parabola and hyperbola. 

We will illustrate by a perspective view (Fig. 309) the 
principle of the relation between the directrix and focus of a 

parabola, and after- 
wards apply it to the 
ellipse and hyperbola. 

ACB is a cone, and 
DEGF the cutting plane. 
H is a sphere touching 
the cutting plane, and 
inscribed in the upper 
portion of the cone. A 
line drawn from c, the 
centre of the sphere, 
perpendicular to the 
cutting plane, will meet 
it in /, which is the 
focus of the parabola. 
The plane KLNM," 
containing the circle of contact between the sphere and the 
cone, intersects the cutting plane in the line ag^ which .is the 
directrix of the parabola. 

A line joining the centre of the sphere with the circle of 
contact, as ce^ is perpendicular to the side of the cone. 
Compare this figure with Fig. 228 (Plane Geometry). 
Let us now refer to Fig. 307. cfe is the inscribed sphere, /is 
the point of contact with the cutting plane, ce is perpendicular 
to the side of the cone, and e determines the level of the plane 
containing the circle of contact. A horizontal line drawn 
through e till it meets the cutting plane produced in d will 
determine the position of the directrix. 

Draw a line from y" perpendicular to the cutting plane till it 




SECTIONS OF SOLIDS. 



179 



meets the line D'E' in /' : this is one of the foci of the elhpse, 
A hne drawn from d perpendicular to the cutting plane will 
determine the directrix ag. 

The line D'E' is the major axis of the ellipse, and if we 
bisect this line by another at right angles to it, and obtain the 
position of the points K and L in the same manner as we 
determined the points b"b'\ KL will be the minor axis. We can 
obtain the other focus and directrix by setting off their distances 
on the opposite side of KL ; or we could construct another 
sphere in the lower part of the cone, and obtain them as 
already described. 

In Fig. y:>'j^ the same construction as previously described 
will determine the position of the directrix and focus ; and as it 
bears corresponding letters, the student should have no diffi- 
culty in understanding it. Compare Fig. 307 with Fig. 309. 

The same thing applies to Fig. 308. 



227. To project the section of a rig-lit vertical cylinder ; the cutting 
plane to be inclined at angle of 36' with the H.P. and 73' with 
the V.P., hut not intersecting the base. This section is an 
ellipse. Construct the sectional area on the H.P. Fig. 310. 

Project the cylinder 
(Prob. 203). 

Find the traces of 
the cutting plane (Prob. 
218). 

Assume a vertical 
plane passing through 
the axis of the cylinder, 
and perpendicular to the 
cutting plane. Draw ab 
perpendicular to the H.T. 
Draw a projector to a till 
it meets the v.T. in a . 
Draw a projector to b till 
it meets XY in b' . Join 
cib'. ab is the H.T. of this 
V. plane, and ctb' the line in which it intersects the cutting plane. 

Draw a projector to o till it meets ab' in o' . 




i8o GEOMETRICAL DRAWING AND DESIGN. 

Divide the plan by diameters into eight equal parts, one 
of these diameters, dh, being in the H.T. of the V. plane. 
Projectors to dh will give d\ h', two of the points in the section. 
Produce the diagonal ie till it meets the H.T. of the cutting 
plane in 7n. Draw the projector in7n\ and draw a line from in' 
through d. This will give the corresponding points z\ e\ in 
the elevation. 

Produce the diameter jf till it meets XY in fi. Draw a 
perpendicular to XY at ft till it meets the V.T. in n'. Draw a 
line from Jt' through d. This will give the corresponding points 
/',/' in the elevation. Find the points c'^g* in the same manner. 

Draw a fair curve through these points for the elevation 
of the section. 

Any number of points in the curve could be found in the 
same manner by drawing additional diameters to the plan, 
but eight points are generally deemed sufficient. 

To construct the sectional area. Find the points B, H, F, D, 
(Prob. 221), and complete the ellipse (Prob. 1 57, Plane Geometry). 

228. To project the section of a sphere ; the cutting plane to be 
inclined at an angle of 35' with the H.P., and 74' with the V.P. 
Fig. 311. 

Find the traces of the cutting plane (Prob. 218). 

As a sphere is a continuous surface without any edges or 
angles, it will be necessary to assume certain fixed lines upon its 
surface in order to determine where the cutting plane will inter- 
sect it ; meridians and parallels are best suited for this purpose. 

Project the sphere with meridians and parallels (Prob. 207). 
It will be better to arrange the meridians on plan so that one 
of them is parallel to the H.T. 

Assume a V.P. perpendicular to the cutting plane and con- 
taining the axis of a sphere. Let ab be the H.T. of this plane, 
and db' the line in which it intersects the cutting plane. Where 
this line intersects the axis will determine <?', and where it 
intersects the meridian dd' will give two points in the section. 

Produce the meridian jk on plan till it meets the H.T. in ?t. 
Draw the projector im', and draw a line from 7t\ through o\ 
till it meets the meridian j'k'. These are two more points in 
the section. 



SECTIONS OF SOLIDS. 



iSi 



Produce the meridian ef till it meets XY in /. Draw a 
perpendicular at / till it meets the V.T. in /'. Draw a line 
from /', through o\ till it meets the meridian f e\ giving two 
more points in the section. Obtain the points h' and g' in the 
same manner, and draw a fair curve through all the points found. 




Fig. 3". 



Drop projectors from each of these points till they meet the 
corresponding meridians on plan, and draw a fair curve through 
them. 

These projections are ellipses. They could also be found by 
first projecting their conjugate diameters, and then completing 
them as in Problem 157. 

The true shape of the section is of course a circle. To obtain 
its radius, bisect cd in j, and draw a line tii through s parallel 
to the H.T. ; st is the radius required ; tu is the major axis, and 
/r^the minor axis of the ellipse. 



I82 



GEOMETRICAL DRAWING AND DESIGN. 



EXERCISES. 



1. Draw the plan of a right pyramid 2j" high, base an equilateral 
triangle of 2" side. The pyramid stands on its base, and the upper part 
is cut off by a horizontal plane i" above the base. Indicate the section 
clearly by light shading. (April, '96.) 

2. An elevation is given of a regular hexagonal prism with its bases 
horizontal (Fig 312). Draw its plan. Also show the true form of the 
section made by the inclined plane shown at bn. (April, '00.) 





Fig. 312 



Fig. 314- 



3. The diagram (Fig. 313) shows the end elevation of a right 
prism if" long with square base, and a horizontal plane Im cutting the 
prism. Draw a plan of the portion below Im. (April, '98. ) 

4. The diagram shows (Fig. 314) the plan of a portion of a sphere. 
Draw an elevation upon the given line xy. (June, '98.) 

5. Plan and sectional elevation are given 
(Fig. 315) of a short length of moulding which 
has been cut across for " mitreing " by a vertical 
plane shown in plan at ////. Determine the true 
form of the section. (June, '97.) 

-^ 6. A right square pyramid, edge of base i|", 
height 2", is cut by a plane which contains one 
edge of the base, and is inclined at 45° to the 
plane of the base. Draw the plan of the section, 
and if you can its true form. (June, '99. ) 

7. A sphere of \" radius has a portion cut off 
by a horizontal plane g" above its centre, and 
another portion by a vertical plane passing ^" 
from the centre. Draw a plan of what remains 
Fig. 315- of the sphere. The section must be clearly 

indicated by lightly shading it. (June, '97. ) 

8. A right cone, height 2^" , radius of base i", stands with its base on 
the horizontal plane. A sphere of i" radius rests on the horizontal 
plane and touches the cone. Draw a plan and elevation of the two 
solids, showing clearly your construction for finding the centre of the 
sphere. Show also the true form of the section of each solid by a 
horizontal plane ^" above the centre of the sphere. (April, '02.) 

i^For fm-ther Exercises see p. 228. ) 




PART III. 



DESIGN. 



CHAPTER XVI. 

The eye and the ear are both pleasantly affected by regularity 
of effect. The ear treats as music the even beats of the air, and 
in the same way the eye is a very accurate judge of evenness 
and regularity. It is strange how unevenness distresses the eye ; 
for instance if the hnes of this page were unequal distances 
apart, the effect would be very irritating. 

But there is a further parallel between music and de- 
coration. A discord has an effect on the enr which may be 
described as unsatisfying, and the "Resolution of the Discord" 
is called for to satisfy the demand for evenness, which has 
been temporarily violated. 

In the same way, in decoration, the eye demands regularity, 
even uniformity of treatment ; while at the same time it has a 
feeling that unevenness or the interruption of uniformity is 
pleasurable, if supplemented by regularity in the whole design. 
The two precisely similar spires of Cologne or Coutances are 
not altogether satisfying, because of their want of contrast ; 
the eye is not challenged by contrast into attention to detail. 
This constitutes the special beauty of the twin, though dis- 
;similar, spires of Lichfield or St. L6, where the eye is piqued by 
the discord into the discovery of the prevailing harmony of 
■design. 

These principles should guide us in decorative design. 
Regularity must prevail in the main, with the unexpected to 
afford and challenge interest. 



i84 GEOMETRICAL DRAWING AND DESIGN. 

Construction Lines on "w^hich Patterns are 
arranged. 

To cover a wide space with regularity of form is an ancient 
device of man ; and in fact the old Greeks were better versed 
in its secrets than are we of this age. And yet, if we note 
their methods carefully, we may observe that geometrical 
balance was the secret of their success. To map out a sur- 
face with a geometrical network, and to fill it with balanced 
forms satisfying to the eye, seemed to them second nature. 
Slowly we are regaining the same instinct, and the beautiful 
decorations of modern artists are the product of dihgent study 
of geometrical design. 











































































1 


1 


























Fig. 316. 



Fig. 317. 





Fig. 318. 



Fig. 319. 



A net is the simplest aid to mapping out a surface, and we 
have in Fig. 316 this groundwork of design, a square net like 
a tennis net ; and in Fig. 317 a 45° net, rectangular, and each 
hne making equal angles of 45° with the horizon. Fig. 318 is a 
60° net, covering the surface with a vertical diamond or lozenge : 
Fig. 319 a 30° net, forming a horizontal diamond or lozenge. 
This mapping out of the whole surface by a network, or skeleton, 
must be the first step, as it is the geometric basis for the 
pattern, the unit of which will be the subject of discussion later. 



DESIGN. 



i8s 



The geometric framework must be prepared before the pattern 
which it is to carry. It will be noticed that in these nets certain 
lines are emphasized, showing how the simple net may be the 
basis of a more complicated network. A square net with 



xxix 


>< 


M>^>^jx 


y 


><i 


X!xp^ 


Vy/ 


W\-y\;^Ay' 


\^ 


V 


XP\D\ 


Y 




vvv 


>sp\5s 


^K'Os 


XXx 


>$><p<. 


X 


44^ 




>^A>\ 


X 


xtxp-!-^.l 


xxx 


X* 


XN\ 


^i^x 


XiXX 


X 


xlxx 


xlxlx 




Fig. 320. 



Fig. 321. 








Fig. 322. 



Fig. 323. Fig. 324. 



1 








1 






















1 


■ jrn. 




















- 







1 1 




n 




1 1 


n 


- -i_ 







Fig. 326. 



Fig. 327. 



diagonals is shown in Fig. 320, and m Fig. 



^,21 a 



45 net 



crossed by vertical and horizontal lines, which form octagons 
and other figures In both Figs. 320 and 321 a square net and 
a 45° net are combined and certain hues are selected. In Fig. 
322, a 60° net with horizontals forms a network of equilateral 
triangles or hexagons. Fig. 323 is a square lattice, and Fig. 
324 a square plaid ; Fig. 325 a lattice of 45° ; Fig. 326 a double 



1 86 



GEOMETRICAL DRAWING AND DESIGN. 



square lattice, and Fig. 327 a square framework. In these we 
see how the choice of Hnes in a square net fills a whole area 
with a skeleton, a process which is simply exemplified in 



Fig. 328. 



X 



m 



X 



m 



X 



m 



Fig. 329. 



1/ 


\ 






xxxx 


A 




X 


/ 


\\, 


X 






A. 




X 


X 


^ 


/; 






X 




X 


/ 


^ 


X 







7 


^ 


\ 


s 


XXX 


\ 






/ 


9>. 


\ 


\^ 


/ 


/ 




4 




\ 








/ 


K TV 



Fig. 330. 



Fig. 331. 





Fig. 332. 



Fig. 333- 



Fig. 328. Fig. 329 shows a skeleton formed by a square and 
diagonals, forming interlacing octagons ; Fig. 330, a roof tiling 
made by the choice of diagonals on a square net ; Fig. 331 a 
diaper constructed in the same way. Fig. 332 is a triangular 



DESIGN. 



187 



framework formed on a 60° net ; the upper part is a drop 
triangular framework, the triangles are only blackened to define 
the triangles of the design ; the lower part shows a series of 
interlacing triangles, identified by their blackened centres. 
Fig. 333 is a hexagonal framework on a 60^ net, with verticals. 
There are three small hexagons to each triangular form. The 
network, shown in the lower right-hand corner, being omitted, 
the lines left form an elaborated skeleton of lines, chosen 
regularly and uniformly related to one another. 




^ 



Fig- 335- 



^ Z S Z \ 




/ \ / \ / 


1 


S z-s_z s 




V i;/' ^ ^ 


1 11 



000 



moo 



Fig. 337- 



Fig. 334 is a framework of interlacing triangles forming a 
star and outlining a hexagon. Fig. 335 shows a framework 
in which interlaced stars all fill up the pattern and give a 
hexagon form ; Fig. 336, a framework of unequal sided octagons, 
formed on a square net. In Fig. 337 a framework of squares 
and stars is formed on a square net which is shown in one of 
them ; the diagonals of the net construct the stars, part only of 
the diagonals being shown. Figs. 338, 339 and 340 show the 
development of simple line patterns, suitable for band ornament, 
formed on a network. In Fig. 341 the selection of lines shows 



i88 



GEOMETRICAL DRAWING AND DESIGN. 



the development of a pattern with the network omitted, and 
Figs. 342, 343 and 344, the selection of the rectangular frame- 
work and added diagonals. Fig. 345 is a plaited band on a 
60° net. 





Fig. 338. 



Fig. 330. 



M-M"i"n 



IIIIII'MI — 

Fig. 340. 



Fig. 341- Fig. 342. 



/__y: 



Fig. 344- 



Fig. 343- 



AAAAAAA7 

Fig. 345. 




Thus far the skeleton or framework has been composed of 
lines selected from a network or diagonals of the net, and formed 
of straight lines only. It remains to add the circle to the 
materials of construction and to proceed to form frameworks 
with its aid. 

Fig. 346. Semicircles are worked on the square net giving 
scale work or imbrication, and Fig. 347 shows a scale work on 
a 30° net. 



DESIGN. 



189 



Fig. 348 shows a framework on a 45° net formed of interlacing 
circles. 

Fig. 349 suggests how, on a 60° network with horizontals, a 
pattern can be developed from interlacing circles and straight 
lines outlining a hexagon. 



^^ 


sz 


"O" 


^^ 


^p 


K 


^^^^ 


^Xm 


» 


^ 


\ 




Fig. 351 



Fig. 349- 




Fig. 352. 





Fig. 35 





Fig. 356. 



Figs. 350-353 give frameworks of circles on a square net 
the net of construction being indicated by dotted lines. 

Fig. 354 shows the change which takes place in the framework 
by working with circles on a 45° net. 

Fig. 355 is a skeleton developed from selected semicircles on 
a square net, and Fig. 356 the same skeleton with additional 
semicircles which cover the vacant intervals. 



I go 



GEOMETRICAL DRAWING AND DESIGN. 



Fig. 357 is a complete skeleton ot interlacing circles and is a 
pure circular pattern. This pattern is based on a 60° net, and 
it is an example of a simple pattern which a child with a pair 
of compasses could draw as he fills a paper with circles. 




Fig. 358. 



^: 


X^ 




J 


V 






J 


x^x 


^ 






\ 


^ 




y 






J 


V 




J\^^ ) 




^ 


C 






~\ 


r ^^^''"'^ 



Fig. 359- 



Fig. 360. 



Fig. 358 is the result of selecting interlacing circles based on 
the 30° net and forms a beautiful geometrical pattern. 

Fig. 359 is a scale work on a square net, the radii of the circles 
being less than the side of the square. 

Fig. 360 is an ogee skeleton and Fig. 361 part of a framework, 
both resulting from the selection of semicircles on a square 
net. 

In Figs. 362 and 363 the selection of quadrants on a square 
net construct two very different skeletons of interlacing ogee 
forms. 



DESTGN. 



Fig. 364 is a skeleton composed of circles and semicircles on 
a square net. 

Figs. 365-367 show an effect of combining circles and straight 




7- 


5<< 


^ ^ ^ 


QXy^ 


^ 2^ 


± 



Fig. 361. 



Fig. 362. 








^j^^j^^j^ 




¥ 


TTT^ 







Fig. 363. 



Fig. 364. 




Fig. 365- 



Fig. 366. 



Fig. 367- 



lines on a square net ; the net being omitted, a framework or 
skeleton remains to receive the details of ornament. 

Figs. 368-371 show the effect of working on a square net, with 
circles, semicircles and straight lines. 



192 



GEOMETRICAL DRAWING A.ND DESIGN. 



Fig. 372 is a double scale work formed by concentric semi- 
circles on a square net. 




Fig. 368. 



i» 


/ 




-A 










V \r 



Fig. 369. 



< 




^ 




V 


xx>c 


^ 


A 



Y 


y 


v^ 


/^ 


^ 


X 










^1 


V 


V 


r 


-N 


r 


^ 


— 





























Fig. 370. 



Fig. 371- 



1 


^ 




X?ft 


L 


^^ 


^ 


J 


"^Yr y/r ^^ 




r 


■N^ 




-.yrvL. jYv^ yr 


^ 


^ 


1- 


^A 


^r V:^ v^ 




372. 



F.g. 373 



Fig. 373 is a framework or skeleton border on a square 
net. 

Fig. 374 IS a framework of quadrants and semicircles with 
diagonals on a square net. 



DESIGN. 



193 



Fig. 375 is a simple roll border formed by parallel lines and 
concentric circles on a 45° net. 




Fig. 374- 



Fig. 375. 




Fig. 376. 





Fig. 377- 



Fig. 378. 




Fig. 379. 



Fig. 380. 



Fig. 381. 



Units of Pattern. 

Having shown how to map out a surface by a network or 
skeleton, in varied ways, we must now proceed to discuss the 
units of pattern with which the surface is to be covered. Figs. 
376, 377 and 378 are simple circular designs, the construction of 
which is obvious ; in Fig. 376, the circles are joined by freehand, 
forming a ball-flower ornament which occurs frequently in 
' Decorated ' Gothic architecture. Fig. 379 illustrates the varia- 
tion of pattern caused by emphasising different parts of a 
design ; the two halves of this unit are really different patterns, 
though the lines in both are the same. Fig. 380 is a triangular 



194 



GEOMETRICAL DRAWING AND DESIGN. 



unit ; for its construction refer to Problem 'j']^ page 49, a triangle 
drawn to touch the dotted circle at CEG gives 
the outline. Fig. 381 is an ogee unit constructed 
on a square and its diagonals. Fig. 382 is a 
square unit, and Figs. 383, 384, units of inter- 
secting squares. Fig. 385 is a unit constructed 




z L^^z: 






Fig. 383. 



Fig. 38A. 





Fig. 386. 



Fig. 387. 





Fig. 391. 
Fig. 389- 

on a 3 by 2 rectangle ; Figs. 386, 387, 388 are square 
units formed by a combination of circles and straight lines. 
Fig. 389 is a square lozenge, enclosing intersecting circles ; 
Fig. 390 consists of concentric squares and circles, all enclosed 
in an equilateral octagon. Fig. 391 is a star formed by two 
interlacing equilateral triangles, with parallel but not equidistant 



DESIGN. 



'95 



sides ; Fig. 392 is a square star unit ; Fig. 393, a unit of interlacing 
squares, enclosing a circular design. Fig. 394 is a star hexagon 
unit. In Fig. 395 geometrical design and freehand are combined 
on an octagon base. 





Fig. 393- 




Fig. 394- 



Fig. 395- 



These few examples of geometrical units of pattern must 
suffice to show the principles of construction. Naturally enough, 
the scheme of decorative treatment usually leaves the trammels 
of geometrical design, relying upon the forms of flowers and 
foliage for the ideas which the repeating patterns carry out. 
But the object in view has been attained if the student has been 
led to see, under the intricacies of decoration, the geometrical 
basis on which it is constructed. 



{Fo7- Exercises see p. 229.) 



196 



GEOMETRICAL DRAWING AND DESIGN. 



The Spacing of Walls and other Surfaces. 

The wall of a modern dwelling-house is usually divided, for 
decorative purposes, into the cornice, frieze, field, dado and 
skirting, as shown in Fig. 396. 




Skirting 



Fig. 396. 



The various methods for covering such given spaces with 
ornamentation by means of geometrical patterns are briefly 
indicated in this section. For the technique of distributing and 
repeating patterns, suitable for such special spaces, in any 
decorative scheme, the student should consult any of the various 
books on Decorative Design. 



DESIGN. 197 

This chapter aims at teaching him how to construct patterns 
geometrically : it is quite another and a larger subject which 
must be separately studied how to make geometrical treatment 
subservient to the decorator's art. 

tmnnj 'gfHTEl -nSZJ 



Fig. 397. Fig. 398. Fig. 399- 



mm. 



Fig. 400. Fig. 401. Fig. 402. 



Fig. 403. Fig. 404. 



JjOOQinki 




Fig. 405. Fig. 406. 




Fig. 407. 

(1). Bands and Borders. 

A wall is very often divided into two parts by a horizontal 
border above the dado ; the treatment of such a border is a simple 
introduction to wall decoration. We begin with a set of bands 
called "Greek Frets"; these are represented in Figs. 397 to 405. 
They are formed by selecting lines from a square net ; an 
introduction was made to this in Figs. 338, 339, 340, where the 
lines of construction are seen. In Fig. 401, a raking pattern, the 



198 GEOMETRICAL DRAWING AND DESIGN. 

diagonals of the net are utilized for construction. Figs. 406-408 
are intersecting frets, 406 being a Moorish plaited band and 407 
an Italian interlacement band ; 408 is a chain band, and 409 a 



^ [7 



^ 



U- 



£} 



Fig. 40S. 




ig. 409. 




Fig. 418. 



straight line band suggesting a series of Maltese crosses formed 
of onyxes. 

With Fig. 410 a series of circular band ornaments is intro- 
duced, leading up to the wave-form introduced in Fig. 414. 



DESIGN, 



199 



The wave and roll form is developed in Figs. 415 to 418, 
and its construction carried on progressively to the complicated 
rolls, Figs. 419, 420. Fig. 421 shows one of the effects of a 




Fig. 419. 




Fig. 420. 




Fig. 421, 



double roll combination suggesting to the student what a 
variety of interesting designs may be evolved from the con- 
structions thus built up. The term "Guilloche" is generally 
apphed to such rolls as are shown in Figs. 419-421. 



200 GEOMETRICAL DRAWING AND DESIGN. 

With Figs. 422-424 is commenced a series of bands having for 
foundation the combination of straight and curved lines, leading 



t>^^ I iF^^^rx 



Fig. 422. 



Fig. 423. 



Ml 10 



Fig. 424. 




Fig. 425. 



Fig. 426. 



i::z: 



1^ 



Fig. 427. 



Fig. 



Fig. 429. 



^_ 



Fig. 430. 



V.-^ 



■^ X 



Fig. 431- 



Fig- 434- 



^p»n^ ^ 




Fig. 432. 



Fig. 433- 



up to Figs. 425, 426, in which the effect of such combinations 
in decoration is exemplified. 



DESIGN. 



Spiral elements are introduced in Figs. 427 to 430, and the 
wave line in Fig. 431 ; and with these materials the beautiful 
bands given in Figs. 432 to 439 are built up. Figs. 432, 433 
and 434 are Greek paintings on terra-cotta. Fig. 435 is a 





I'"!^'- 435- 



Fig. 436. 





Fig. 437- 



Fig. 43S. 




Fig. 439- 

French mural painting of the 13th century; Fig. 436, a border 
from a picture by Domenicino (i6th century). Fig. 437 is a 
Greek terra-cotta, and 438 an Early Gothic French ornament. 
Fig. 439 is an "aesthetic" design. The two "repeats" in these 
cases are a sufficient guide to the complete scheme. 



{For Exeirises sec p. 231.) 



202 GEOMETRICAL DRAWING AND DESIGN. 

(2). Defined Areas— Walls, Cer-ings, Floors, etc. 

We come now to deal with the treatment of certain defined 
areas ; hitherto we have only dealt with schemes meant for 
general areas. Fig. 440 indicates the simplest division of a 
square so as to give border and corners, while Fig. 441 only 




Fig. 440. 



Fig. 441. 






Fig. 442. 



Fig. 443- 



Fig. 444. 





Fig. 445- 



Fig. 446. 



affords a plain border, inside which a Maltese cross divides up 
the space for further ornamentation. Fig. 442 shows a circular 
centre, and Fig. 443, circular corners. Fig. 444 is a simple indi- 
cation of the centre and corners due to the inscribed circle, while 
Fig. 445 shows an elaboration of the centre with wide borders. 
Fig. 446 is a square design, with a centre which arrests the 



DESIGN. 



203 



eye by being unexpected, as the arms are not radial. Fig. 447 
is a square ceiling design ; the top outer circle is completed to 
show the construction, the true design being given in the lower 
part. Fig. 448 is a panelling for a ceiling from a tomb in Rome, 
forming, as in all the cases we have been considering, a skeleton 





n ^ / 

DUOL_ 


D 

D 



Fig. 447. 




Fig. 448. 




Fig. 449. 



Fig. 450. 



Fig. 451. 



XX 



Fig. 452. 



i_j 



\2^^n 




Fig- 453- 



for decorative treatment. Fig. 449 is a rectangular space which 
is divided up by selected lines of a square net, with certain 
diagonals. Fig. 450 is a square lozenge, separating the corners 
of the rectangle. Fig. 451 is a cross of St. Andrew, with 
circular centre, affording triangular panels. Fig. 452 shows a 
rectangular moulding with subdivision of the space in the 



204 



GEOMETRICAL DRAWING AND DESIGN. 



form of a cross. Fig. 453 shows a simple circular treatment 
of a panel ; Figs. 454, 455, two lunettes with circular treatment. 
Fig. 456 is a simple lunette and spandrels. Fig, 457 gives a 
subdivision of a circular space suggesting tracery, and Fig. 
458 a trefoil treatment. Fig. 459 gives the complete hexagonal 
system of circles, inscribed in the given area, from which the 
artist may select or emphasize symmetric arcs, so as to produce 





Fig. 454- 



Fig. 455- 




Fig. 456. 




Fig. 457 



Fig. 459- 



Fig. 460. 



an extraordinary variety of design. Fig. 460 is a tracery design, 
worked from the hexagon of the outer circle. Fig. 461 is the 
subdivision of an octagonal area, the figure being built on a 
square net ; Fig. 462 on the 45° net. Fig. 463 is a star figure, 
which is built up inside an octagon. This is a sample of the 
many dififerent stars which may be formed by varying the radius 
of the dotted circle of construction. The student will find a 



DESIGN 



205 



useful exercise in making several examples, which will illustrate 
the various designs which result from the change of this circle. 
For example, if the radius of the dotted circle be made about 
half of that in Fig. 463, and radii of the larger circle be drawn 
to the points of the star, we have the familiar appearance of the 
old mariner's compass as it used to be before the advent of 
the spider web of the Thomson compass. 





Fig. 46 



Fig. 462. 




Fig. 463. 





Fig. 464. 



Fig. 466. 



Fig. 464 again is a sample of the division of a hexagonal 
space which is suggestive of many varieties. The elementary 
feature, the joining of all the points of a hexagon, suggests in 
appearance the outlines of a transparent icosahedron : a star, 
inscribed in a circle is formed in the hexagon. The figure itself 
can be amplified by further outlining the details, the simple 
plan adopted in the figure is to follow each line with another 
line parallel to it at a fixed distance throughout. It is evidently 
possible to produce a large variety of divisions of the hexagon 
on this model, and the introduction of circular arcs will add still 



2o6 GEOMETRICAL DRAWING AND 3i:: 



more. L 5, 466 the space treated is an equilateral 

triangle ; here a. simple plan is to divide it either by a central 
hexagon, as in Fig. 465, or by a central circle, as in Fig. 466. 
In each of the examples, th« feature of doubled and parallel lines 
is introduced. 

The object which is airied at in presenting these figures is 
to suggest modes of setting out defined areas — walls, ceilings, 
and floors — for decorative treatment. It is this general 
delineation which is the particular aim of geometrical design. 
In the completed scheme, no doubt, the framework or scaffolding 
will be entirely lost, but the aim of the designer is to afford a 
pleasing arrangement of the space available ; and in such a 
way that the eye should not be arrested by the ornamentation 
forming a simple network. 

It has often been observed that a wall paper for a sick room 
must not offend in this particular. If a patient's eye is con- 
tinually challenged by a repeat in network over the surface, or 
a figure pattern too prominently recurrent over a large area, the 
mental disturbance is serious. The mind falls to counting the 
patterns in wea:risome persistence, and the more the mind is 
beyond physical control, the more serious is the effect. 

Hence it is desirable that the designer should practise the 
division of spaces by forms which have a pleasing intricacy of 
geometrical balance. The methods here indicated are only 
specimens which should suggest the lines which may be 
followed, and it is hoped that they will lead the student to 
exercise his ingenuity in planning more serious designs. 



[^For Exercises see p. 232.) 



TION. 



ORNAMENTATION. 



CHAPTER XVII. 

Lettering. 

It is very important for the student to be able to letter his 
drawings well. The design of Roman lettering is a serious 
study of which it is possible only to give the merest outline 
in Figs. 467, 468. The basis of construction is a square of 



TX 



^ 




Fig. 467. 



Fig. 468. 



rectangular net, with circles to guide in the formation of the 
serif The revival of the old style of lettering, with the serif 
inclined as shown in Fig. 467, letter L, has given a new and 
interesting impetus to the artistic study of lettering. 

These two examples are taken from Albert Durer's Geometrica^ 
in which he gives methods for drawing Roman capitals. 
Suppose as a groundwork a square. The thick strokes 
are \ of the square and the thin strokes are ^^ of the 
square. The serifs are constructed on circles of j 
diameter. Mr. Walter Crane says : " Letters may be ...ken 
as the simplest form of definition by means of line. They 
have been reduced through centuries of use from their pi' nillve 
hieroglyphic forms to their present arbitrary and fixed vpes : 
though even these fixed types are subject to the Vitriation 
produced by changes in taste and fancy." 



2o8 GEO "LTRICAl^' DRAWING AND imSOJ^. 



Shields. 

Shields often appear as an element of ornamentation, and 
must be treated in accordance with the rules of heraldry. It is 
a very early and general rule that metal must not be placed 
upon metal, nor colour upon colour ; but that they must be 
placed in contrast. ' Or,' gold, and ' argent^ silver, are the 
metals used, and ' aziire^ blue ; gules^ red ; ^ purpurea ' vert.^ 
and ' sable ' or black are the colours usually employed. 

Black may be taken here as indicating metals and white as 
colours. Furs, erynme and vair^ are also used ; but of these 
ornamentation takes no account. Figs. 469 to 472 show the 
principal divisions of shields. Fig. 469, checquy^ a shield divided 



Fig. 469. 



Fig. 470. 



Fig. 471, 



Fig. 472. 



in chequers or small squares like a chessboard ; the number 
varies. Fig. 470, quarterly^ the field being divided into four 
quarters. Fig. 471, the Pale or a vertical strip set upright in the 
middle of the shield and one-third of its breadth. Fig, 472 
represents a band division. The upper part is the chiefs occu- 
pying one-third of the height, the fcss is the bar, horizontally 
placed in the middle of the field. 

Among the other divisions of the field must be reckoned the 
chevron^ a A -shaped strip and the cross, usually a Greek cross of 
equal arms. When plain, this cross is in breadth one-third of 
the shield ; but its varieties are manifold. 



Diaper, Chequer, Spot, Powder. 

ct diaper is a repeated pattern covering a given surface 
intervals ; for example. Fig. 473, which gives the appear- 



an 






encaustic tiling with no variation of pattern ; Fig. 474 
imple of a chequer^ which consists of a repeat alternately 



vviih. v<.v:ant spaces. 



ORNAME/TATIQTxN. 



209 



Fig. 475 is 2iSpot pattern ; it must be observed that the spot 
pattern has large regular vacant intervals, the chequer, a vacant 
space equal to that of the repeat. 



it 



Fig. 475- 



<> 



^ 



O < 



Fig. 476. 



la 



EI 



13 



S 



a 



lara 



HSia 



H 



Fig. 477. 



Fig. 478. 





Fig. 479. 



Fig. 480. 



Powder again differs from spot in point of scale, the unit of 
powdering should be small and simple, . Powdering consists of 
small and insignificant units of repetition and may be combined 
with spot. For example, Fig. 476 is a combination of spot 
and powdering. 

Fig. 477 is a stripe and band pattern. Units of pattern when 
arranged in narrow lines are called ' stripe,' and when wider, 
'band.' For instance, Fig. 478 is the elementary stripe and 
band, the simplest form of this decoration. Fig. 479 is a 

O 



2IO GEOMETRICAL' i5^RAWING AND DESIGN. 

chequered band, with stripes arranged for diagonal decoration ; 
Fig. 480 is a paneUing, derived from bands, and Fig. 481 a piece 
of parquet flooring. Fig. 482 is an application of diaper or 
chequer in more elaborate form. It represents an inlaid work 
of independent, interlacing squares. In Fig. 483, crosses are 




Fig. 4S1. 



Fig. 482. 




Fig. 4?3- 



Fig. 484. 



arranged as a diaper for mosaic decoration. Fig. 484 is a chequer 
formed by a combination of Greek fret and square foliage ; Fig. 
485 is a more elaborated diaper or spot pattern on a circular 
basis and constructed on a 45° net, which forms part of the 
design. Fig. 486 is a diaper pattern formed in marble mosaic, 
from San Vitale, Ravenna. Fig. 487, a ceiling panelling 



ORNAMENTATION. 




Fig. 4S5. 



wm 




^ 



m 



u 




ilL 



.^ 




P^P^P 



1 



nr 



^ 



i: 



^ 



Fig. 486. 




Fig. 487. 



212 GEOMETRICAL DRAWING AND DESIGN. 

conveying the impression of spot and powdering. It is taken 
from a mediceval enamel in Cologne. Fig. 488 is a mosaic 



%M' 


^ 


((1^)) 


(TA ) (?•) T 



Fig. 48E 



flooring of simple construction and a charming intricacy. 
Fig. 489 is one panel of ceiling decoration, in which only one 




Fig. 489. 



repeat is shown. Fig. 490, a scale-work diaper, a simple 
construction if based on a 45° net. Fig. 491, a very pleasing 
combination of square and circular treatment suitable for a 
mosaic flooring. Fig. 492 is a reproduction of a Byzantine 
bas-relief from the Cathedral of San Marco, Venice. It is 



ORNAMENTATION. 




Fig. 490. 




Fig. 491- 



r 




— 


^ 


^^^^^ 


^ 



Fig. 492. 



214 



GEOMETRICAL DRAWING AND DESIGN. 



a beautiful example of varied interlacing. The constructions 
are extremely simple, and the efifect is due to the unexpected, 




Fig. 493. 

which piques the interest. Fig. 493 is an Egyptian ceiHng 
decoration ; in reality it is a simple diaper and spot with spiral 



■^^-^^^"^ 



r 




Fig. 494. 



construction. Fig. 494 is the corner of a rectangular mosaic. 
Fig. 495 is an interlacement band ornamenting a northern MS. 
of the eighth or ninth century ; Fig. 496, a geometrical band 
decoration in coloured marble from the wall of Mackworth 
Church, Derbyshire. Fig. 497, a band of Moorish mosaic from 



ORNAMENTATION. 



215 



Granada, which, it may be observed, is interchangeable, the 
black and white spaces being exactly equivalent. 




Fig. 495- 

Decoration based on geometrical construction is of infinite 
variety, and a thorough familiarity with geometrical relations 
will regulate genius and inspire the designer with a just view of 




Fig. 496. 



^^ 



Fig. 497. 

those forms which will be pleasing and restful to the eye, as well 
as satisfying to the natural demand for relief. 



{For Exercises see p. 233.) 



ARCHITECTURE. 



CHAPTER XVIII. 



Arch Forms and Tracery, 

The semicircle was the first arch-form to appear in building. 
In the earhest times (Nippur, B.C. 4000), such arches were 
employed below the level of the ground. The tendency of the 
round arch to sink when bearing any weight led these early 
builders only to use it with that strong lateral support. The 
Romans employed it largely ; as their bridges and aqueducts 
standing to this day exemplify. The Pont du Gard, near 
Nimes (B.C. 19), is about 160 ft. high and 880 ft. long ; built of 





cJeryii circular 

Fig. 498. 



Fig. 499. 



Fig. 500. 



large stones without cement, it affords a striking example of the 
stability of the semicircle. 

The round arch may be semicircular, as Fig. 498 ; segmental, 
as Fig. 499 ; "elliptic" or three-centred, as Fig. 500. It maybe 
"stilted," as Fig. 501, the semicircle being continued in straight 



ARCHITECTURE. 217 



lines ; or a " horse-shoe," as Fig. 502, the circle itself being 
continued. This last is a Moorish feature, the semicircular a 
Roman, and the stilted arch a Byzantine feature. 

The Basilica, or Hall, originally a place of business, became 
in the hands of the Roman builders a structure of heavy round 



Fig. 501. Fig. 502. 

arches and circular windows. This form was adopted for the 
early Christian churches ; we have such a building at Brixworth, 
Northants. 

The Norman builders (1066 to 1190) followed up the semi- 
circular brick arches, building them in stone ; the heavy round 
pillars with cushioned capitals, and the heavy arches with 
geometrical ornamentation being the successors of the Roman 
work. 

The origin of the pointed arch has been much discussed ; an 
example is found in Cairo, a horse-shoe arch of the ninth 
century ; in fact, at that time, this form was regarded by the 
Moslems as their special emblem. It is met with in the 
Crusaders' churches throughout the twelfth century. If in Eng- 
land it has been developed from Norman arcading, as is often 
supposed, there is an example of the process on the towers of 
Southwell Minster. There is a semicircular arcading ; then on 
another face, semicircular arcades intersect ; the lancets thus 
formed become windows in the next, and, finally, the arcading 
disappears and lancet-headed windows are seen alone. 

The Transitional Period, heavy Norman work with pointed 
arches (about 1140 to 1200), is well exemplified in the arcading 
of St. John's Church, Chester ; best, however, in the Choir of 
Canterbury Cathedral. 

The Early English, or Lancet Style (about 1190 to 1300), was 
thus developed at the end of the twelfth century from the 
circular or Romanesque ; the lancet windows (Fig. 503) and the 
clustered pillars (Fig. 543) giving elegance and lightness. 
Characteristic ornament was introduced, and a style grew up 
which is essentially English. 



2lJ 



GEOMETRICAL DRAWING AND DESIGN. 



At this time also Tracery begins to appear, apparently 
developed as follows. Lancet windows are placed together ; in 
Salisbury Cathedral, our great instance of a complete Early 
English building, are seen combinations of two to even seven 
lancet windows together. Again, three windows are often placed 





Fig. 504. 



together, the middle window being higher than the other two. 
The three would be treated as one window, one arch moulding 
including them all. Again, two lancet windows are placed close 
together with a trefoil or quatrefoil (Fig. 512) above them, and 
are treated as one window ; in Westminster Abbey, a small 
triangle appears besides the small quatrefoil. This "wall 
tracery " led to " plate tracery," where the wall is thinned to a 





Fig. 506. 



single piece of stone ; and plate tracery led to "bar tracery," 
where the wall between windows becomes a bar, and the 
quatrefoil becomes a geometrical design in the arch. 



ARCHITECTURE. 



219 



This leads to the Geometrical Period (about 1260 to 1320), 
and, at this point, the study of tracery begins. The window is 
divided, according to its size, by a number of vertical bars or 
"mullions" ; the arch is equilateral (Fig. 504) and filled with 
circles, trefoils and curved triangles in strictly geometrical 
design. The " element " of the design (Fig. 509) is the skeleton 
showing the centres of the circles of the tracery. The element 
of some windows is nothing but a 45° network, with quatrefoils 
as a unit of pattern. The element (Fig. 509) will form an 
interesting exercise for the student to complete, and the ruined 
window at the end of the book (p. 262) will be found a more 
difficult problem of the same sort. In the Geometrical Period, 
circular foils (Fig. 512) alone are used ; in the last quarter of the 
thirteenth century, pointed foils (Fig. 513) are introduced. 

The Decorated Period (about 1300 to 1375). The simplicity 
of style, so far described, led, in the hands of ingenious designers, 
to flowing curves, such as Figs. 511, 515, 518 suggest. This 
flowing style is a purely English development ; it may be called 
Flowing Decorated (about 131 5 to 1360). The "ogee" arch 
(Fig. 507) and the pointed foiled arch are introduced ; also 





Fig. 508. 



bands of wavy foliage of a natural design. But this leads us 
beyond the scope of Geometrical Design. 

It seems, however, that the excess of ornament led to a return 
of simpler forms in the last quarter of the fourteenth century, 
and introduced the Perpendicular Style (about 1400 to 1545)- 
This deserves our attention because of its geometrical character. 



GEOMETRICAL DRAWING AND DESIGN. 



Straight lines are the feature of this style in place of the flowing 
lines of the later Decorative Period, circular lines and circular 
cusping. Horizontal transoms are introduced into the large 
windows at this period, and become a decorative feature. The 



I 


1/ 


% 












Fig. 510. 




P^ig. 509. 



Fig. 51 



enclosing square which is used for construction in the element 
509 becomes the outline of door and window. The four-centred 
arch is due to this period (Fig. 508). 

This, again, is a style which is peculiarly English, and is 
more appreciated nowadays. Perhaps it is too mechanical, and 





Fig. 512. 



Fig. 513- 



there may be too much repetition of ornament ; but to this 
period belong the great East Anglian churches, and such 
masterpieces as the Tower, Choir, and Lady Chapel of Gloucester 



ARCHITECTURE. 



221 



Cathedral. It forms a grand close to the development of our 
English Church Architecture. 




Fig- 517- 



Fig. 518. 



Greek and Roman Mouldings. 

Mouldings have been called the alphabet of architecture ; 
they are the elements which determine and give expression 
to the parts of a building. 

The simple forms of moulding are : 

The Roman ovolo, a quadrant (Fig. 519). 



S 



C 



Fig. 519- 



Fig. 520. 



Fig. 521. 



The cavetto, or hollow (Fig. 520), which is the reverse of 
the ovolo. 

The torus, or half-round (Fig. 521). 



222 GEOMETRICAL DRAWING AND DESIGN. 

Other mouldings and cornices or other designs can be made 
by arranging these with flat spaces, above, below, or between 
them. 

A fillet (Fig. 522) is a small flat face, and the torus when 
small is called a bead or astragal (Fig. 523), which may be 



Fig. 522. Fig. 523. Fig. 524. 

incised, so as not to project from the flat surface. Several 
parallel beads together are called reeding. 

The cyma recta (Fig. 524) is formed by combining ovolo 
and cavetto, the hollow being uppermost, and is suitable for 
a cornice. 

The cyma reversa, or ogee (Fig. 525), a similar combination, 
with the hollow at the bottom, is suitable for a base 
moulding. 

The scotia is formed by two quadrants, as shown in Fig. 526. 



Fig. 525- 




Fig. 526. Fig. 527 



The Greek mouldings correspond with these, but their 
section is not circular like the Roman. They^ are for the 
most part constructed with conic sections, viz. ellipses, or 
parabolas. But in all probability they were drawn in by hand. 

Some construction lines for these mouldings are shown in the 
figures, 527, ovolo; 528, cyma recta; 529, cyma reversa or 
ogee ; 530, scotia. 



ARCHITECTURE. 



The Bird's Beak or Hawk's Beak moulding (Fig. 531) is 
common in Greek Doric architecture. It is a cyma recta 
surmounted by a heavy ovolo which casts a bold clear shadow 




Fig. 529. 



Fig. 530. 



Fig- 531 



over the cyma. It is particularly interesting, because it dis- 
appears from architecture entirely after the best period of 
Athenian art. 

Greek architecture is distinguished by the grace and beauty 
of its mouldings ; it has been remarked that their sections 
are mostly elliptic. They are, however, not regular curves ; 
they must be drawn, rules cannot be given for describing 
them. Symmetry, proportion, and refinement are the charac- 
teristics of Greek ornament. 

The mouldings of Roman origin are in general form the 
same as the Grecian, but their contour is bolder and section 
circular. 

The ornamentation of Roman mouldings was no doubt 
borrowed from Greece, but it is less restrained. Roman 
architecture is overdone with ornamentation ; foliage and 
various subjects in relief covering every moulding and surface. 




Fig- 532- 



Fig- 533- 



One or two characteristic specimens of ornamentation may 
be given. Fig. 532 is the egg and tongue or arrow on the 
Greek echinus or ovolo ; and Fig. 533 the egg and dart on a 



224 



GEOMETRICAL DRAWING AND DESIGN. 



Roman ovolo. A bead or astragal may be divided up, 

as Fig. 534. Figs. 535 and 536 -^ \/^\/- — V^ 

are typical ornamentations of torus V TV /\ f\ J 
moulding. Fig. 534- 





Fig. 535- Fig. 536. 

Fig. 537 is a Greek leaf ornament for cyma reversa from 
the Erechtheum at Athens, and Fig. 538 a Roman leaf ornament 
for the same type of moulding. 





Fig. 537- Fig. 538. 

Gothic Mouldings. 

Mouldings were developed contemporaneously with the other 
features of Gothic architecture. In the Norman period, as might 
be expected, these are square and circular in section. 

The stock Norman moulding (Fig. 539) consists of a broad 
hollow surmounted by a broad fillet, from which it is cut off by 
a small sunk channel ; in fact, the hollow is set off by ' quirks ' or 
returns. 



Fig. 539- 




Fig. 540. 



A plain round projection is frequently found with a narrow 
fillet above it, the quirks being chamfered (Fig. 540). This 
moulding is called a bowtel. 



ARCHITECTURE. 



225 



There are very deep mouldings over the round Norman 
doorways, consisting simply of squares and circles, at Iffley, 
Oxford, for example (Fig. 541). 





Fig. 541- 



Fig. 542. 



The Early English architects developed this style of mould- 
ing, retaining circular forms and almost entirely eschewing ogee 
or reversed curves. A single specimen only need be given, 
part of a doorway, at Woodford, Northants (Fig. 542). 

It may be noticed here that fillets are freely run down the 
face of circular mouldings, that each circle is defined, and that 
there is no returned curve. These may be taken as the simple 
characteristics of Gothic mouldings. 



Gothic Piers. 

One architectural feature which should be mentioned is the 

©development of the simple circular 
and square pier of the Norman 
style into the elaborate piers of the 
Perpendicular period. 

The massive Norman pier in the 
hands of the early Enghsh builders 
was lessened in size and had shafts 
set round it, as Fig. 543, a specimen 

Ofrom Salisbury Cathedral or the 
north transept of Westminster 
Abbey. The shafts increased in 

Fig. 54^. 

number and were incorporated in 
the body of the pier, still preserving a circular contour. 

p 




zb 



GEOMETRICAL DRAWING AND DESIGN. 



The contour became of a lozenge plan in the style called 
Decorated, and the number of pillars is much increased. The 
shafts were arranged diamond wise, so many as would stand 
close together, with only a fillet or small hollow between them. 




i:Z\ 




Fig- 544- 



Fig- 545- 



Fig. 544 shows a Decorated pier at Dorchester, Oxfordshire, 
and Fig. 545 a Perpendicular pier from Rushden, Northants. 



{For Exercises see p. 235.) 



MISCELLANEOUS EXERCISES 

FROM EXAMINATION PAPERS OF THE 
BOARD OF EDUCATION 



ORTHOGRAPHIC PROJECTION (Ch. XIV.). 

1. A plan and elevation are given of a l)uUress projecting from a 
wall (Q. i). Draw a fresh elevation on a vertical plane which makes 
an angle of 45° with the plane of the wall. 






Q.I. 

2. The end elevation is given (Q. 2) of a small cofter or caddy, the 
length of which is to be 3". The lid has four sloping faces, which all 
make the same angle (30°) with the horizontal. Draw the plan of the 
lid. 



228 



GEOMETRICAL DRAWING AND DESIGN. 



3. The plan is given (Q. 3) of an octagonal tray or dish, the height 
of which is |". Make an elevation on the given xy. Only the visible 
lines need be shown in elevation, and the thickness of the material is 
to be neglected. 

4. The diagram (Q. 4) represents a doorway in a wall, the door 
being shown opened at an angle of 45° with the surface of the wall. 
Draw an elevation of the door when closed, t'.e. showing the true form 
of the panels. On/j' the door need be drawn, not the surrounding 
mouldings. Your construction 'uusi be shown. 




Q-5- 



5, The plan is given (Q. 5) of a square slab, a5 being one edge of a 
square base. Draw an elevation on a vertical plane at right angles to 
ad. Show in plan the section made by a horizontal plane containing 
ad. 










Q. 6. 



I. — .....J 



Q.7. 



6. The plan and end elevation are given (Q. 6) of a simple hut. 
Draw an elevation on a vertical plane w^hich makes an angle of 30" 
with the long walls of the hut. (N.B. — Only the visible lines need be 
shown. ) 

7. The diagram (Q. 7) shows an elevation and section of an opening 
in a wall. Make a second elevation when the face of the wall makes 
an angle of 45° with the vertical plane of projection. 



MISCELLANEOUS EXERCISES. 



229 



8. The diagram (Q. 8) shows an elevation of a square slab, Ji B 
being one side of a square face, 3" long. Draw the plan of the slab, 
and show a cylindrical hole of 2" diam. pierced through its centre. 






9. The plan is given (Q. 9) of a square prism, of which AB represents 
a square face, 3" each side. Determine the elevation of the prism on 
the given xy, and add the elevation of a circular hole of 2" diameter 
piercing the centre of the prism. 

10, The diagram (Q. 10) shows a perspective sketch of part of a 
buttress. Make an approximate sketch plan and a side elevation. 



SECTIONS OF SOLIDS (Ch. XV.). 

1. Make an approximate sketch plan, and also a sectional elevation, 
of the mortar of which a perspective sketch is given (Q. i;, assuming 





that the inside form is a hemisphere. Show clearly any construction 
you would suggest. The size ot your drawing should be about 3 times 
that of the diagram. 



230 GEOMETRICAL DRAWING AND DESIGN. 



2. A right cylinder of 2" diameter is cut by a plane, making an 
angle of 30° with the axis of the cylinder. Show the true form of the 
section. 

3. The plan is given (Q. 3) of a right cone (diameter of base, 2^"), 
of which y is the vertex. Make an elevation on the given xy, and show 
the section by a vertical plane parallel to the base and i" from it. 



DESIGN. 

CONSTRUCTION LINES AND UNITS, pp. 184-195. 

1. Sketch four illustrations of ornament formed by circle?, and 
explain the object they fulfil in certain cases. 

2. Draw clearly, with instruments or freehand, the geometrical basis 
on which the given "diaper" pattern (Q. 2) is constructed. Plan the 
scale of your diagram to show two "repeats" of the pattern in a 
width of 3^". Show 5 or 6 repeats in all. (Only sufficient of the 
ornamental detail need be sketched to indicate its position.) 





Q. 2. Q. 3 

3. Sketch, with instruments or freehand, 07ie unit of the given 
diaper (Q. 3), showing clearly your method of setting out its details. 
Make your drawing about three times as large as the diagram. [The 
height and width of the figure are equal.] 

4. Draw, freehand or with instruments, a system of construction 
lines on which the given system of quatrefoils (Q. 4) can be built up. 
Show how you would determine the centres of the arcs, and the points 
of contact. Three or four repeats of the unit should be indicated about 
twice the size of the diagram. 



MISCELLANEOUS EXERCISES. 



5. Show clearly any geometrical construction you would think useful 
in setting out the given "Tudor rose" (Q. 5) about four times the 
dimensions of the diagram. You need only sketch so much of the 
flower as is needed to illustrate your construction. 





cj. 4. Q. 5- 

6. Draw, with instruments or freehand, the system of construction, 
lines on which you would build up the given repeating pattern (Q. 6). 
Show also one complete unit of the repeat. 





Q. 6. 



Q. 7. 



7. Show clearly how you would set about drawing the given pattern 
(Q. 7). Mark what you consider the unit, and indicate any construc- 
tions you think needful. Only two complete adjacent units need be 
shown, about four times the scale of the diagram. 

8. Any triangle can be repeated so as to cover a space without leaving 
interspaces. Show how this can be done, and sketch four other shapes, 
rectilinear or curved, which will repeat in a similar way. 

9. A floor has to be covered with tiles which are squares and regular 
octagons in shape. Sketch the pattern so formed, showing clearly how 
you would set it out, and marking the unit of repeat. 



232 



GEOMETRICAL DRAWING AND DESIGN. 



SPACING OF SURFACES, pp. 196-201. 

1. Draw a Greek fret without keys, but with a border at top and 
bottom ; the fret, borders and spaces, to be |" wide. Also another fret 
with a tee and a border top and bottom ; borders, spaces and fret each 
\" wide. Explain the principles of the ornamentation and the surfaces 
to which they can be applied. 

2. Draw three sorts of frets, and another fret on the slant, and an 
instance of frets alternating with ornamented panels, each i" high. 
Explain the principles and the surfaces to which they are properly 
applied. 

3. Draw a band ornamented with square panels filled in with some 
usual Greek ornament, divided from one another by Greek keyed frets, 
meeting at centre line of panel, with border top and bottom : height 
exclusive of border i", length 3|-", to contain three panels. Explain the 
principles of the ornamentation. 



Q.5. 




4. Make a drawing of one unit of the given border (Q. 4), increasing 
the length of the "unit" to 2\", and the other dimensions in proportion. 
Indicate the method by which the figure should be constructed. 



5. Make an enlarged copy of one unit of repeat of the given border 
(Q. 5), the height of your drawing being increased to iA^"and the length 
in proportion. Show a construction for obtaining the divisions of the 
circle. 

6. The diagram (Q. 6) shows half a circular plate. Indicate a 
method by which the leading wave line of the ornament could be made 
up of arcs of circles of equal radius. 

Your drawing should be three times the diagram. 



MISCELLANEOUS EXERCISES. 



233 



7. Show the construction you would employ in setting out the given 
border (Q, 7). 

Not more than three repeats should be drawn. 




Q. 7. 



DEFINED AREAS, pp. 202-206. 

1. Show the construction lines upon which the ornament in the 
square panel (Q. i) has been designed. 

2. The figure (Q. 2) shows one quarter of the decoration of a circular 
plaque. Complete the circle and set out the panels. The position only 
of the freehand ornament need be indicated. 





Q.I. 



Q. 2. 





3. Show how you would proceed to modify the given figure (Q. 3) so 
as to make the central panel a regular octagon, the width of the four 
side panels remaining unchanged. 

4. What is the geometrical basis of the given design (Q. 4) for chip- 
carving? (N.B. — Show only the main lines. Do not attempt to copy 
the figure completely.) 



234 GEOMETRICAL DRAWING AND DESIGN. 



5. Draw a coffered ceiling, 6" square, with beams round the outside, 
with a circular panel in the middle, and four angle or spandrel panels, 
the spacing to be in harmonic proportion. Ornament the coffers and 
the soffits of the beams if you can, and explain the principles of the 
ornamentation. 



ORNAMENTATION, pp. 207-215. 

1. Draw, with instruments or freehand, two different arrangements 
by which the two given ornaments (Q. i) may be used, alternating with 
one another, so as to form a ' •' diaper " pattern. 

(The ornaments may be roughly sketched simply to indicate their 
position. ) 

2. Show how, by repeating and reversing the given lines (Q. 2), an 
*' all over " pattern may be obtained. Indicate the lines of construction. 








Q- I. Q. 2. Q. 3. Q. 5. 

3. Sketch, with instruments or freehand, an "all over" diaper 
pattern formed by repeating the given unit (Q. 3). Show nine repeats 
of the unit, with the leading lines of the construction. Make each unit 
about three times the size of that in the diagram. (The height and 
width of the figure are equal. ) 

4. The outline of the diaper pattern formed by placing repeats of the 
given figure (Q. 3) in contact with one another is made up of semicircles. 
Draw at least four repeats of the outline so as to show clearly where 
centres and points of contact of the semicircles occur. Make each unit 
about five times the size of that in the diagram. 

5. The diagram (Q. 5) represents a stencilled ornament which it is 
desired to repeat so as to form a "diaper" pattern. Sketch two ways 
in which this can be done, the repeats of the unit being placed adjacent 
to one another. Show four repeats in each case about twice the size of 
the diagram. 

6. Indicate clearly a geometrical basis for the given repeating pattern 
(Q. 6), and show what you consider to be the unit. (The freehand 
ornament should only be shown once.) 



MISCELLANEOUS EXERCISES. 



235 



7. It is desired to restore the complete circular ornament of which a 
fragment is shown (Q. 7). How would you do this? 

The restorations of the dark portions should be disregarded altogether 
in your drawing. 




g. 8. 



8. Indicate any geometrical construction you think desirable in setting 
out the given pattern (Q. 8). 

Show clearly what you consider the unit of the pattern. (N.B. — Do 
not try to copy all the details ; only show enough to make your meaning 
clear.) 

9. Draw, with instruments, specimens of scale work (imbrication), 
showing two scales and a half to each in length, each scale to be %" 
wide. Give specimens of scales formed of half circles, oblongs with 
rounded ends, outlines of leaves, leaves with an outer margin and ribs, 
and one whose leaves are double ogees with the point of the leaf turned 
up, and with ribs, and one of trefoils with margin and filled with 
ornament. State to what surfaces they can be properly applied. 



236 



GEOMETRICAL DRAWING AND DESIGN. 



ARCHITECTURE, pp. 216-225. 

1. Draw the given diagram of window tracery (Q. i), using the 
figured dimensions. The arch is " equilateral." 





^■f'-^S- 




Q. 2. 



Q-3- 



2. Draw the given outline of window tracery (Q. 2), using the 
figured dimensions. The arch is "equilateral" and all the arcs are of 
equal radius. 

3. Draw the "cyma recta" moulding shown (Q. 3), adhering to 
the given dimensions. The curve is composed of two quarter- circles 
of equal radii, tangential to one another and to the lines AB and CD 
respectively. 

4. Draw the "scotia" moulding shown (Q. 4). The curve is 
made up of two quarter-circles of i" and ^" radius respectively. 






Q. 5- 



O. 6. 



5. Draw the 

dimensions. 



"rosette" shown (Q. 5), according to the figured 



6. Draw the "ogee" arch shown (Q. 6) to a scale of 2' to i". 
The arcs are all of 2' radius. The methods of finding the centres and 
points of contact must be clearly shown, 

7. Draw the moulding shown (Q. 7), adhering strictly to the 
figured dimensions. The arc of V radius is a quadrant. 

8. Diaw the " cyma recta " moulding shown in the diagram 
(Q. 8), using the figured dimensions. The curve is composed of two 
equal tangential arcs each of f" radius. 



MISCELLANEOUS EXERCISES. 



237 



9. Copy the cornice given (Q. 9), increasing the total height to 2|" 
and the other measurements in proportion. "\'ou may draw the "cyma" 
moulding by any geometrical construction that seems to you suitable. 




Q. 7. 





■■m'--> 




1 






/^ 


r '^ 


J 


\ 


Q. 8. 


^ 






Y 












Q. 9. 




10. How would you draw the leading lines of the window tracery 
given (Q. 10)? (N. B. — Do not try to copy the whole figure, but only 
clearly indicate your method.) 

11. Show how to set out the given figure (Q. 11). 

12. Make an approximate sketch plan or elevation (but not both) of 
the given column base (Q. 12). 



EXAMINATION PAPERS IN GEOMETRICAL 
DRAWING. 

BOARD OF EDUCATION. 

GENERAL INSTRUCTIONS TO CANDIDATES. 

You may not attempt more than five questions, of which three only 
may be chosen from Section A, and two only from Section B. But no 
award will be made to a Candidate unless he qualifies in both sections. 

All your drawings must be made on the single sheet of drawing paper 
supplied, for no second sheet will b& allowed. You may use both sides 
of the paper. 

None of the drawings need be inked in. 

Put the number of the question close to your workings of problems, 
in large distinct figures. 

The value attached to each question is shown in brackets after the 
question. 

A single accent (') signifies_/^^/ ; a double accent (") inches. 

Questions marked (*) have accompanying diagrams. 

Your name may be written only upon the numbered slip attached to 
your drawing paper. 

I. 

SECTION A. 

In this section you may attempt three questions only, 

SHOWING your knowledge BY THE USE OF INSTRUMENTS. 

The constructions must therefore be strictly geometrical, and not the 
result of calculation or trial. 

All lines used in the constructions niust be clearly shown. 

Set squares may be used wherever convenient. Lines may be bisected 
by trial. 



EXAMINATION PAPERS. 



239 



1. Draw a diagonal scale yV of full size, by which feet, inches, and 
eighths of an inch may be measured up to 5'. 

By means of this scale construct a triangle having its altitude 2' 2J", 
one side 2' 5I-", and base 2' o|". Write down the length of the third 
side. (20) 

*2. Draw the given figure by inscribing seven equal squares within 
a circle of i|" radius. (20) 





Q.3- 




Q. 6. 



Q. 7. 



*3. Copy the given guilloche ornament, making the radii of the 
circles |", ^", |", and i" respectively. (16) 

4. Describe two circles of f" radius touching each other, and one of 
i" radius touching the first two. Then describe a fourth circle touching 
all these three. All constructions must be clearly shown, and all 
contacts indicated. (20) 

5. A quadrilateral ABCD is to be described about a circle of |" 
radius. AB = 2%", AD=2h,", and the angle BAD is 42°. Draw the 
figure, show all points of contact, and write down in degrees the angle 
ADC. Then draw a similar quadrilateral having the radius of its 
inscribed circle i|". (18) 



240 GEOMETRICAL DRAWING AND DESIGN. 

*6. An elevation is given of part of an octagonal pillar with square 
base. Draw its plan, and an elevation on a vertical plane which makes 
30° with one of the vertical faces of the base. (20) 

*7. The diagram shows the elevation of a solid composed of a 
cylinder capped by a portion of a sphere. Draw the true form of the 
section made by the plane indicated by the dotted line. (20) 



SECTION B. 
In this section you may attempt two questions only. 

All freehand work employed in this section must be neat and 
careful, and its intention must be made quite clear. 
All constructions must be clearly shown. 

*8. Redraw the given pattern, with instruments, altering the pro- 
portions so as to make the octagons regular and of |" side. (20) 




Q. 8. 



*9. Show the geometrical framework on which the given pattern is 
based. Show clearly what you consider the unit of repeat, and draw 
four repeats, sketching only enough of the pattern to make your 
intention plain. (18) 

*10. Draw the geometrical constructions you think necessary in 
setting out the plate, making your drawing double the dimensions of 
the print. It will suffice if one quarter of the design is clearly shown. 

(20) 

"■■"11. Draw, using instruments, one of the cusped arch-forms, pre- 
serving, as accurately as you can, the proportions of arch and cusping. 
Make your drawing twice the dimensions of the print. (20) 



EXAMINATION PAPERS. 



241 



*12. Make an approximate sketch plan and front and side elevations 
of the flower-holder, using instruments where you think advisable. 
Arrange your drawings so that one is projected from another. (20) 




Q. 9. 



242 GEOMETRICAL DRAWING AND DESIGN. 




Q. II. 



EXAMINATION PAPERS. 243 




Q. 12. 



244 



GEOMETRICAL DRAWING AND DESIGN. 



II. 
SECTION A. 

1. A drawing made to a scale of ^ of full size has to be re-drawn so 
that the dimensions shall be enlarged by one-fifth. Make a scale for the 
new drawing, showing feet (up to 3'), inches, and (diagonally) eighths of 
an inch. 

Figure the scale properly, and show by two small marks on it how 
you would take off a distance of i' 4§". (22) 




g. 2. Q. 5. 

■^2, Copy the given figure, making the diameter of the outer circle 
38". Show how to determine all the points of contact between circles 
and straight lines. {20) 





Q. 6. 



Q. 7. 



3. Two points, A and B, are i" apart. Find a third point, C, I "6" 
from A and 2" from B. With centre A and radius AB describe a circle. 
Describe a second circle touching the first at B, and passing through C. 
Describe a third circle of i" radius touching the first two (but not at B). 

(18) 

4. The foci of an ellipse are 3" apart. A point Z>, on the curve, is i" 
from one focus and 3" from the other. Draw the ellipse and a normal 
to it at D. (16) 

*5. The point of intersection of the two given lines being inaccessible, 
draw through the point Z' (i) a line which would pass through the point 



EXAMINATION PAPERS. 



245 



of intersection of the two given lines, and (2) a line making equal angles 
with the two given lines. (16) 

*6. The diagram shows the plan of a right square pyramid. Draw an 
elevation. (18) 

■^7. The elevation is given of an "elbow" formed by two pieces of 
cylindrical piping. Draw the plan, and also the true form of the inter- 
section of the pipes. The thickness of the material may be neglected. 

(22) 



SECTION B. 

*8. Show how you would set out a geometrical framework for the 
given pattern, so as to exhibit a number of repeats. Only enough of the 




246 



GEOMETRICAL DRAWING AND DESIGN. 



ornament should be sketched to show quite plainly what you consider 
the unit of repeat. (18) 

9. Regular pentagons will not by themselves cover a surface. Draw 
any form which in combination with regular pentagons would serve this 
purpose. Make a diagram of the pattern formed, and show clearly what 
you would use in practice as the unit of repeat. .' (lb) 




*10. Show what geometrical aids you would employ in drawing the 
square panel given. Make your drawing about the size of the figure. 

(18) 

■^11. What geometrical means would you use in setting out the fan 
shown? Make your drawing about the size of the figure. (16) 

*12. Draw a plan of the given table, showing clearly any construc- 
tions you would use. (18) 



EXAMINATION PAPERS. 



247 



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248 



GEOMETRICAL DRAWING AND DESIGN. 



III. 
SECTION A. 

1. Make 2. plain scale, to show feet and inches up to 5 feet, on which 
a distance of 3' 6" is represented by 4I". Finish and figure the scale 
neatly and carefully. 

Draw to this scale an oblong 3' 3" x 2' 5", and in the centre of it 
place a second oblong of the same shape but having its longer sides 
2' 9". Measure the breadth of this smaller oblong to the nearest half- 
inch. (20) 





Q. 2. 



Q. 3. 






Q. 4. 



Q. 6. 



Q. 7. 



*2. Complete half \h^ given figure, which is made up of regular 
pentagons. Make the radius of your enclosing circle 2". Show clearly 
any construction you employ. (20) 

*3. Copy the given figure, making the radius of the outer circle if, 
and that of the inner one in proportion. (18) 

*4. Draw the given figure. The curves are to be composed of arcs 
of circles of 05" and 1-5" radii. (20) 



EXAMINATION PAPERS. 



249 



5. Two straight lines, AB and CD, are 3" and 3-5" long respectively. 
A is 1-5" from C, while B is i" from D and 3-5" from C. Describe two 
circles each touching both AB and CD, one passing through A, the 
other through ^. (16) 

*6. The outline of a "scotia" moulding is shown. If two lengths 
of this moulding are "mitred" together at right-angles, show the true 
form of the cut surface of the mitre. (18) 

*7. A lamp-shade, in the form of a truncated regular hexagonal 
pyramid, is made of six pieces of card, of the exact shape and twice the 
size shown. Draw its plan and elevation in any position. (20) 

SECTION B. 

*8. Show, using instruments, how you would set out the geometrical 
framework of the given openwork panel. None of the "cusping" 
need be drawn. (20) 




Q. 8. 

*9. Draw neatly with instruments the framework or "net" of the 
given pattern. Show clearly what you intend to be the unit of repeat, 
and finish not less than four of these so as to show repeats both in 
width and height. (2o) 



250 



GEOMETRICxVL DRAWING AND DESIGN. 



*10, Set out, as nearly as you can, the construction needed for the 
geometrical part of the ornament round the semicircular door-head of 
which about half is shown. Only one unit of each ornament need be 
completed. Make your drawing twice as large as the diagram. (i8) 




Q. 9. 




EXAMINATION PAPERS. 



251 



"^11. Redraw the given figure, altering the proportions of the parts so 
that the sides of the four corner panels shall be exactly half those of the 
centre panel. Make the side of the outside square 3", and the margins 
throughout 02" wide. (16) 



D 


1 I 


D 








D 


1 1 


D 



Q. II. 

'^12. Draw an approximate sketch-plan, with front and side eleva- 
tions, of the given steps. Arrange your drawings to show how one is 
projected from the other. (18) 




252 



GEOMETRICAL DRAWING AND DESIGN. 



IV. 

SECTION A. 

1. Six feet are represented on a drawing by one inch. Make a scale 
for the drawing by which single feet can be measured up to 40', and show 
inches diagonally. 

Figure the scale properly, and show by two small marks on it how you 
would take off a distance of 20' 8". (22) 





■^2. Make a copy of the diagram, using arcs of I J" and |" radii only. 
Show clearly how all points of contact are obtained. (18) 

*3. Copy the diagram, making the radius of the outer circle if". 

(20) 

■^4. The diagram shows a symmetrical figure composed of straight 
lines and five semicircles of equal radii. Draw a similar figure having 
a total height of 3^". (22) 





Q. 4. 



Q. 6. 



5. Construct a regular nonagon of if" side. Describe a circle touch- 
ing all the sides of the nonagon. Within the circle inscribe a regular 
nonagon having its sides parallel to those of the first one. 

If you employ a protractor for measuring an angle, this must be 
clearly shown and the number of degrees stated. (16) 



EXAMINATION PAPERS. 



253 



*6. The diagram shows the elevation of a short prism, or slab, the 
bases of which are eqiiilata-al triangles. Draw the plan, and write 
down the angle which the bases make with the vertical plane of pro- 
jection. (18) 

7. A right cone, diameter of base 3I", height 2^", has the plane of its 
base inclined at 45° to the horizontal plane. Draw the plan of the cone, 
and of its section by a plane parallel to the base and i^" from the vertex. 

(20) 

SECTION B. 

*8. Draw a geometrical framework on which the given diaper pattern 
may be constructed. You are advised to draw the framework geometri- 




Q.8. 



254 



GEOMETRICAL DRAWING AND DESIGN. 



cally, and to sketch, freehand, just enough of the pattern to illustrate 
your meaning. Show very clearly what you consider to be the unit of 
repeat. (i8) 

*9. The diagram shows a pattern composed 
entirely of semicircles. Show two other ways 
of arranging semicircles so as to produce a 
repeating pattern, with the necessary construc- 
tions for determining centres. (i6) 

•^10. Show what geometrical help you would 
use in setting out the plate shown in the 
diagram. Assume that AB, the diameter of 
the octagon, is 5". (16) 

*11. Indicate what geometrical construc- 
tions you would employ in setting out the circular window shown in 
the diagram. The "cusping" need not be shown. (18) 




Q. 9. 




Q. 10. 



■^12. Sketch appro.ximate front and side elevations of the chair shown 
in the diagram. (ij^j 



EXAMINATION PAPERS. 255 




Paper IV. Q. 11. 



256 GEOMETRICAL DRAWING AND DESIGN. 




Paper IV. Q. 12. 



EXAMINATION PAPERS. 



257 



SECTION A. 

*1. The given line AB represents a length of 15 centimetres. Con- 
struct a scale by which decimetres, centimetres, and millimetres can be 
measured up to 3 decimetres. Figure the scale properly, and show by 
two small marks on it how to take off on it a distance of 263 milli- 
metres. 

By means of the scale draw a ciicle of 125 millimetres' radius, and in 
it place a chord 189 millimetres long. Write down in millimetres the 
distance of this chord from the centre of the circle. 

(i metre = 10 decimetres, or 100 centimetres, or looo millimetres.) 

(20) 



B 



2. Construct a regular decagon or llgure of 10 sides, each side being 
1" long. Within it mscnhQ Jive equal circles, each touching two of the 
others and one side of the decagon. 

(N.B. — If a protractor is usecl for measuring an angle, such use must 
be clearly shown.) (20) 

*3. The curve of the given " scotia " moulding is made up of two 

quadrants or quarter-circles. Draw the figure from the given dimensions. 

(N.B.— The diagram is not drawn to scale.) (16) 




Z'4--- 

Q. 3- Q- 4- 

*4. The diagram shows the leading lines of a window composed of 
three semicircular-headed " lights " included under a three-centred arch. 
The side lights are to be each 3' wide, while the middle light is to be 
2' wide and to have the centre of its semicircle 2' above those of the 
other two. Parts of the side semicircles also form part of the enclosing 
arch. Draw the figure to a scale of 2' to 1". Show clearly how you 
obtain all points of contact. (20] 

(N.B. — The diagram is not drawn to scale.) 



258 



GEOMETRICAL DRAWING AND DESIGN. 



6. Two fixed lines AB and CD, of indefinite length, cross one another 
at an angle of 70°. A third line EF, 3" long, is movable, so that the 
end E travels along the line AB while the end F travels along the line 
CD. Draw the complete curve traced by the middle point of EF. 

(It will be sufficient to find some 12 to 16 points on the curve.) 

(18) 

■''6. The diagram gives the front elevation of a regular five-pointed 
star, which is cut out of material 'f thick. Draw the plan of the star, 
and also a second elevation on a vertical plane inclined at 60° to that of 
the given elevation. (20) 




Q. 6. 



*7. An elevation is given of a sphere upon which rests a conical 
lamp-shade. Draw the plan of the shade. (20) 




Q. 7. 



EXAMINATION PAPERS. 



259 



SECTION B. 

*8. Draw with instruments the main geometrical construction lines 
you would employ in setting out the given pattern. Show plainly what 
you consider the practical or working unit of repeat. (20) 




9. Show how you would arrange a number of circular discs of |" 
radius as a diaper pattern — 

(i) When each disc touches four others, 
(ii) When each disc touches six others, 
(iii) When each disc touches three others. 
Draw the necessary constructional framework in each case, and show 
clearly what you consider the unit of repeat in each pattern. (20 marks). 



26o GEOMETRICAL DRAWING AND DESIGN. 

_ *10. Set out carefully the lines of one quarter of the given book- 
binding, making your drawing to twice the scale of the diagram. 




■•■11. Make a modified version of the border of interlacing circles 
which the radii shall be ^", g", §", and i" respectively. Show at 'l( 
two repeats of the unit. 




EXAMINATION TAPERS. 



261 



*12. Sketch approximately, in outline, the plan and the front and 
side elevations of the workbox and its lid shown in the diagram, 
omitting all merely ornamental details. Arrange your drawings so as 
to show how one is projected from another. (20) 




Q, .2. 



262 GEOMETRICAL DRAWING AND DESIGN. 



Complete the design of this ruintd window, following the indicj 
of every fragment \\ hich remains. 




Date Due i 


'^^(.Y 2 ^ -S? 








QP- -^ 








DEC 2? 


1993 






















































X 








































































Library Bureau Cat. No. 1137 



WiELlS BINDERV 
ALTHAM, MASS. 
JAN. 1948 



NC715.S7 




^ 3, 5002 00047 5942 

Spanton, J. Humphrey ^v^-r«_ 

Geometrical drawing and design. 



SCffi